Describing fiber products in stable $\infty$-categories
In fact what you need is that your ∞-category is additive (i.e. that it has direct sums and that the canonical commutative monoid structure on the mapping spaces is group-like). All stable categories are additive, so I'll just prove it in this case.
Step 1: The pullback $X\times_Z Y$ is equivalent to the pullback $(X\times Y)\times_{Z\times Z}Z$. This fact is true in every ∞-category with finite limits
I'm sure there are many ways of proving this. One way of doing it (thanks to Omar Antolín-Camarena for the idea) is to consider the diagram
$$\require{AMScd} \begin{CD} Y @>>> Z @<<< Z\\ @VVV @VVV @VVV\\ \ast @>>> \ast @<<< Z\\ @AAA @AAA @AAA\\ X @>>> Z @<<< Z \end{CD}\,,$$
where $\ast$ is the terminal object. Taking the limit first in the horizontal direction and the in the vertical direction we obtain $X\times_Z Y$, while taking the limits in the opposite order we obtain $(X\times Y)\times_{Z\times Z}Z$. Since both methods compute the limit of the whole diagram, the two objects must be equivalent.
Step 2: The diagonal $Z\to Z\oplus Z$ is the fiber of the "difference" map $Z\oplus Z\xrightarrow{(1_Z,-1_Z)} Z$
This is where we use the additivity hypothesis. Recall that one way of phrasing that the ∞-category is additive is that the shear map $$Z\oplus Z\xrightarrow{\left(\substack{1\, 1\\ 0\, 1}\right)} Z\oplus Z$$ is an equivalence. Then there is an equivalence of sequences $$\require{AMScd} \begin{CD} Z @>{\left(\substack{0\\1}\right)}>> Z\oplus Z @>{(1\,0)}>> Z\\ @| @V{\left(\substack{1\,1\\0\,1}\right)}VV @|\\ Z @>{\left(\substack{1\\1}\right)}>> Z\oplus Z @>{(1\,-1)}>> Z \end{CD}$$ Since the top one is a fiber sequence, so is the bottom one.
Step 3 Profit!
Let us consider the following diagram $$\require{AMScd} \begin{CD} X\times_Z Y @>>> Z @>>> 0\\ @VVV @V{\left(\substack{1\\1}\right)}VV @VVV\\ X\oplus Y @>{(f\,g)}>> Z\oplus Z @>{(1\,-1)}>> Z \end{CD}$$ Since the left and right squares are cartesian (by step 1 and 2), so is the big one, which was the thesis.
Bonus question: In every $\infty$-category with finite limits, if you have a square of pointed objects $$\require{AMScd} \begin{CD} F @>{a}>> X\\ @V{b}VV @V{f}VV\\ Y @>{g}>> Z \end{CD} $$ there is a canonical equivalence between the fiber of the map $F\to X\times_Z Y$ and the fiber of the induced map $\mathrm{fib}(a)→\mathrm{fib}(g)$. This can be proven as before, by considering the diagram $$\require{AMScd} \begin{CD} \ast @>>> X @<<{a}< F\\ @VVV @V{f}VV @V{b}VV\\ \ast @>>> Z @<<{g}< Y\\ @AAA @A{g}AA @AA{1_Y}A\\ \ast @>>> Y @<{1_Y}<< Y \end{CD}\,,$$ and taking the limit in the two directions (you can find more details in the first section of this paper). Incidentally this object is called the total fiber of the square.
Since in a stable ∞-category a map is an equivalence iff the fiber is trivial, this gives an affermative answer to your query.
A proof of this in the context of stable derivators can be found in my paper Mayer-Vietoris sequences in stable derivators with Moritz Groth and Kate Ponto. It is rather more verbose than Denis's, but one reason for that is that it's intended to be fully precise. In an $\infty$-category, of course, you can't actually just write down some objects and arrows and have a "diagram"; there are coherences that need to be produced too, and kept track of all the way through. Derivators are a way of forcing oneself to remember that. (With that said, it's quite possible that if Denis's proof were made precise it would still be shorter than ours.)