Right Kan extension of $\mathcal{F} : \mathsf{\Delta} \rightarrow \mathsf{Top}$.
It's an odd duck, this right Kan extension, and trivial more often than you might think! Skip to the end of this long answer to see examples.
Call the right Kan extension you're after $R : \mathsf{sSet} \to \mathsf{Top}$. Using the formula for the Kan extension as an end, you get that for any simplicial set $X$, $$R(X) = \int_{[n]\in\mathsf\Delta} |\Delta^n|^{\mathsf{hom}_\mathsf{sSet}(X, \Delta^n)}.$$
Now look at that exponent, $\mathsf{hom}_\mathsf{sSet}(X, \Delta^n)$. The simplicial set $\Delta^n$ is the nerve of the category $[n]$, which is the poset $0 \le 1 \le 2 \le \cdots \le n$ thought of as a category in the usual way. The nerve functor $N : \mathsf{Cat} \to \mathsf{sSet}$ has a left adjoint $\tau_1 : \mathsf{sSet} \to \mathsf{Cat}$ (called the fundamental category functor). So we get a natural bijection $\mathsf{hom}_\mathsf{sSet}(X, \Delta^n) \cong \mathsf{hom}_\mathsf{Cat}(\tau_1 X, [n])$.
This shows that $R(X)$ only depends on $\tau_1X$, say $R = R' \circ \tau_1$, where $R' : \mathsf{Cat} \to \mathsf{Top}$ is the right Kan extension of your functor $\mathcal{F} : \mathsf\Delta \to \mathsf{Top}$ along the functor $\mathsf\Delta \to \mathsf{Cat}$ that sends $[n] \in \mathsf\Delta$ to the category we also called $[n]$. As an end, $$R'(C) = \int_{[n]\in\mathsf\Delta} |\Delta^n|^{\mathsf{hom}_\mathsf{Cat}(C, [n])}.$$
Now another similar round of simplification: this category $[n]$ is a preorder and the inclusion $\mathsf{Pre} \to \mathsf{Cat}$ of preorders into categories has a left adjoint: associated to a category $C$ there is a preorder $C^\mathrm{pre}$ whose elements are the objects of $C$ and where $x \le y$ if there is a morphism $x \to y$ in $C$. We have a natural bijection $\mathsf{hom}_\mathsf{Cat}(C, [n]) \cong \mathsf{hom}_\mathsf{Pre}(C^\mathrm{pre}, [n])$, where that last $[n]$ denotes the obvious poset. Again we can factor $R' = R'' \circ (-)^\mathrm{pre}$ and write $R'''$ as a right Kan extension.
In fact, the inclusion of posets into preorders also has a left adjoint, so we can further simplify! Given a preorder $P$, the relation $\sim$ given by $x \sim y \iff (x \le y \text{ and } y \le x)$ is an equivalence relation and the quotient $P/\!\!\sim$ becomes a partial order by defining $[x] \le [y] \iff x \le y$.
So, all told, we get that $R(X)$ only depends on the poset $P(X):=(\tau_1X)^\mathrm{pre}/\!\!\sim$, and is given by the formula $$R(X) = \int_{[n]\in\mathsf\Delta} |\Delta^n|^{\mathsf{hom}_\mathsf{Poset}(P(X), \Delta^n)}.$$ It's probably a good idea to pause and describe this poset more directly. The elements of the poset are equivalence classes of 0-simplices of $X$, where two 0-simplices $x$ and $y$ are equivalent if there exists a directed path of 1-simplices from $x$ to $y$ and also a directed path of 1-simplices from $y$ to $x$. In the poset we have $[x] \le [y]$ if there exists a directe path of 1-simplices from $x$ to $y$ (this does not depend on the representatives $x$ and $y$). Notice, in particular that $P(X)$ only depends on the 1-skeleton of $X$.
OK, now some examples!
$R(\Delta^n) = |\Delta^n|$, because right Kan extension along a fully faithful functor (here, the Yoneda embedding $y$) really extends the functor $\mathcal{F}$. Since $P(\Delta^n) = [n]$, this means that whenever $P(X) = [n]$, we get $R(X) = |\Delta^n|$. For example any $X$ that is a subsimplicial set of $\Delta^n$ containing the spine (the path $0 \to 1 \to \cdots \to n$) has $R(X) = |\Delta^n|$.
If $X$ has the property that for any $0$-simplices $x$ and $y$ there is a directed path of 1-simplices from $x$ to $y$ (and thus also one from $y$ to $x$, then $P(X)$ has only one element and thus $R(X)$ is a point. Examples of this include any simplcial set with only one 0-simplex, and any connected Kan complex.
Added later: You can simplify that last description down to a finite limit. Given a poset $P$, let $\mathcal{S}_P$ be the category with objects order-preserving surjective functions of the form $P \to [n]$, and morphisms from $\lambda : P \to [m]$ to $\mu : P \to [n]$ be surjective order-preserving maps $\alpha : [m] \to [n]$ with $\alpha \circ \lambda = \mu$. There is a functor $F_P : \mathcal{S}_P \to \mathsf{Top}$ given by $F_P(P \to [n]) = |\Delta^n|$. Then this space we're after is just $\tilde{R}(P) = \lim F_P$.
If $P$ is a finite poset, then this finite limit is a convex subpolytope of a big product of simplices, and is not hard to compute in small examples.
- For an antichain with two points, you get $[0,1]^2$.
- For an antichain with three points, you get $\{(x_1,x_2,x_3,y_1,y_2,y_3) \in [0,1]^6 | x_i \le y_i, x_i \le y_{i+1}\}$ (with indices mod 3).
- For the poset $a<c>b$ (with $a$ and $b$ incomparable), you get $\{(x,y_1,y_2) \in [0,1]^3 | x\le y_i\}$, which is a pyramid with a square base.
I'll try to enlight a little the obscure point. (Co)Ends are very powerful because it allows to make quick computation but it sometimes hides the meat.
The first thing to recall is that every presheaf $P$ on the category $\mathsf C$ can be written as the colimit of the Yoneda embedding over the category of its elements (it is a corollary of Yoneda's lemma): $ \DeclareMathOperator*{\colim}{colim} $ $$ P \simeq \colim_{x:a \to P} y(a) $$
So if $Q$ is a presheaf on $\mathsf D$ that commutes with limits, we have the following: $$\begin{split} \hat{\mathsf C}(P,Q\mathcal F) \simeq \lim_{x:a \to P} \hat{\mathsf C}(y(a),Q\mathcal F) \simeq \lim_{x:a \to P} Q(\mathcal F(a)) \\\simeq Q(\colim_{x:a \to P}\mathcal F(a)) \simeq Q(\operatorname{Lan}_y\mathcal F(P)) \end{split}$$ where the last isomorphism is from the usual formula of left Kan extension.
Now apply that with $Q$ being the functor $\mathsf D(-,d)$ (which commutes with limits), and you end up with: $$ \hat{\mathsf C}(P,N_\mathcal F(d)) \simeq \mathsf D(\operatorname{Lan}_y\mathcal F(P),d)$$
This is more or less exactly the same as the (co)end chain, but I hope it brings some light to people not at ease with (co)end yoga.
Fun facts. Left Kan extensions really are everywhere in this framework:
- The nerver $N_{\mathcal F}$ is the left kan extension of $y$ along $\mathcal F$.
- The fact that every presheaf is a canonical colimit can be express by the following: the identity functor on $\hat{\mathsf C}$ is the left Kan extension of $y$ along $y$.
This can be a clue of why the right Kan extension of $\mathcal F$ does not have good property in general compared to the left one.
Because probably someone else (someone marked it with the favourite "star") might be interested in the answer of the above question I managed to found out an answer partly. So here is an answer (with some details omitted) for the second question.
As I mentioned above and someone can easily suspect, we have a kind of natural adjunction between two very interesting functors, namely $\mathsf{sSet} \overset{|-|}{\underset{S}\rightleftarrows} \mathsf{Top},$ hence as (almost) always there should be a nice categorical pattern behind the stage. In our case we can have the following setup instead of these specific categories. So instead of $\mathsf{\Delta}$, we can have any small category $\mathsf{C}$, instead of $\mathsf{Top}$ we can have any cocomplete, locally small category say $\mathsf{D}$ and suppose that by $ \mathcal{y}: \mathcal{C} \rightarrow \hat{\mathsf{C}},$ we denote the so-called Yoneda Embedding. In other words assume that we have the following diagram $$ \begin{array}{ccc} &&\mathsf{C} & \xrightarrow{\mathcal{F}} & \mathsf{D}\\ &&\mathcal{y} \searrow& & \\ &&& \hat{\mathsf{C}}, &&&& \end{array} $$ Because any left/right Kan extension point-wise can be approximated by certain colimit/limit (of a diagram over a certain comma category, whoever is interested in a detailed description which are going to omit out of simplicity's sake, Emily's Reihl - Category theory in context last chapter contains a thorough proof of the latter) and since by assumption $\mathsf{D}$ is cocomplete, the left Kan extension of $\mathcal{F}$ along $\mathcal{y}$ exists (in this situation sometimes the extension is called Yoneda Extension also). So, as we do with the singularization functor, someone can observe that there is a functor in $\mathsf{D}$ doing always the same job as $S$ does for $\mathsf{Top}$. Indeed, we can just define $$ N_{\mathcal{F}}: \mathsf{D} \rightarrow \hat{\mathsf{C}}, \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d \mapsto \mathsf{hom}_{\mathsf{D}}(\mathcal{F}-,d),$$ check by assumption for $\mathsf{D}$, that this is a well-defined functor to presheaves of $\mathsf{C}$ indeed. Now, there is an obscure point (which the reader can probably sort out by looking out into the aforementioned book too), that the left/right Kan extensions can be written in terms of co/ends. So a standard description of $Lan_{\mathsf{y}}{\mathsf{F}}$ can be given as follows $$ Lan_{\mathsf{y}}{\mathsf{F}} = \int^{c} \mathsf{hom}_{\mathsf{D}}(\mathcal{F}c,-) \thinspace.\mathcal{F}c,$$ where the intermediate dot stands for the coproduct inside $\mathsf{D}$. So keeping all the above in mind and exploit some basic properties (like the contravariant $\mathsf{hom}$ change the coends to ends for instance) of coends we can prove the adjointness.
$$\mathsf{hom}_{\mathsf{D}}(Lan_{\mathsf{y}}\mathcal{F}, d) = \mathsf{hom}_{\mathsf{D}}( \int^{c} \mathsf{hom}_{\hat{\mathsf{C}}}(\mathcal{y}c,P) . \mathcal{F}c,d)=\int_{c}\mathsf{hom}_{\mathsf{D}}( \mathsf{hom}_{\hat{\mathsf{C}}}(\mathcal{y}c,P) . \mathcal{F}c,d)= \int_{c} \mathsf{hom}_{\mathsf{Set}}(\mathsf{hom}_{\hat{\mathsf{C}}}(\mathcal{y}c,P), \mathsf{hom}_{\mathsf{D}}(\mathcal{F}c,d))=\int_{c}\mathsf{hom}_{\mathsf{Set}}(Pc, \mathsf{hom}_{\mathsf{D}}(\mathcal{F}c,d))= \mathsf{hom}_{\hat{\mathsf{C}}}(P,N_{\mathcal{F}}(d))$$ Hence in pure categorical language (rather complex if someone doesn't know the terminology) there is a concrete pattern behind the adjunction.
P.S.
For whoever wants to go through the details of the above proof, can find answers in the following link which I find quite useful in that context https://arxiv.org/ftp/arxiv/papers/1501/1501.02503.pdf . Any recommendations, alternative proofs popping out off the top of your head are more than welcome.