What actually means to compute things at tree level?

Suppose you want to compute a correlation say in Euclidean signature $$ \frac{1}{Z}\int D\phi\ \prod_i \phi(x_i)\ \exp\left(-\frac{1}{\hbar}S(\phi)\right) $$ with $$ S(\phi)=\frac{1}{2}(\phi,A\phi)+g\int dx\ \phi(x)^4 $$ where $(\phi,A\phi)=\int\ dx\ dy\ \phi(x)A(x,y)\phi(y)$ for some "matrix" or rather kernel $A$. One usually does that by expanding the functional integral as $$ \int D\phi\ \prod_i \phi(x_i)\ \exp\left(-\frac{1}{\hbar}S(\phi)\right)= \sum_{N=0}^{\infty} \int D\phi\ e^{-\frac{1}{2}(\phi,\frac{1}{\hbar}A\phi)} \prod_i \phi(x_i)\ \times \left(-\frac{g}{\hbar}\int dx\ \phi(x)^4\right)^N $$ and using the Isserlis-Wick Theorem. The contractions involve the covariance or the inverse of the operator in the free quadratic part $$ C=\left(\frac{1}{\hbar}A\right)^{-1}=\hbar \times A^{-1}\ . $$ So when you count powers of $\hbar$ you get $\hbar^{E-V}$ where $E$ is the number of edges and $V$ is the number of internal vertices. For a connected graph (in the case of a vacuum diagram for simplicity) one has the Euler-Poincaré type relation $$ E-V=L-1 $$ where $L$ is the number of independent loops. So the power of $\hbar$ which appears is $\hbar^{L-1}$ which therefore counts loops. The lowest order is when $L=0$ and $E=V-1$, i.e., when the graph is a tree.

In principle in QFT you want to calculate the interaction picture unitary time evolution operator. Then you can use the operator to evolve quantum states just like you would in normal quantum mechanics. So, for example, suppose you set up your particle accelerator to generate a certain state $|i\rangle$ at a time $t_0$, and then after some time $t_1 - t_0$ has passed you want to know the probability it has evolved into a state $|f\rangle$, you calculate the probability amplitude:

$$ \langle f|U(t_1,t_0)|i \rangle $$

The expression for $U(t_1,t_0)$ is given by Dyson's formula:

$$ U(t_1,t_0) = \mathcal{T}\left\{\exp\left(-i\int_{t_0}^{t_1}H_{\mathrm{int}}(t) \mathrm{d}t\right)\right\} $$

where the $\mathcal{T}$ stands for time ordering and the expansion of the exponential is given by a Dyson series, the $n$th term of which has the form:

$$ U_{n}(t_1,t_0) = (-i)^{n}\int_{t_0}^{t_1}dt\int_{t_0}^{t}dt'\int_{t_0}^{t'}dt'' \cdots\int_{t_0}^{t^{(n-1)}}dt^{(n)}\mathcal{T}\left\{H_{\mathrm{int}}(t)H_{\mathrm{int}}(t') \cdots H_{\mathrm{int}}(t^{(n)})\right\} $$

Typically the Hamiltonian has a coupling constant at the front, and so the Dyson series becomes an expansion in powers of that coupling constant (the power of which is equal to the number of vertices in the corresponding Feynman diagram). I don't really know what he means by powers of $\hbar$.

Typically the above integral is impossible to compute independently like you would in normal QM, but we can use a bit of magic called the LSZ reduction formula to write a probability amplitude like:

$$ \langle f|U(-\infty,\infty)|i \rangle \sim \langle 0|\mathcal{T}\left\{\phi(x)\cdots\phi(x^{(b)})\phi(y)\cdots\phi(y^{(b')})\ U(-\infty,\infty)\right\}|0 \rangle $$

where $|0\rangle$ is the vacuum state of the free theory, $\phi$ are the fields and you have one for each particle in the initial state (labelled with $x$) and one for each in the final state (labelled with $y$). I've left out some integrals and constants that you'll get to eventually but aren't important for this high-level overview. Notice that this only holds in the $\pm\infty$ time limit, which is called the 'adiabatic limit'.

Once you have this, you can substitute in your $n$th order term for $U$ and get the $n$th order approximation for your probability amplitude. Remember that the $H_{\mathrm{int}}$ are typically products of fields themselves, so substituting just gives a larger product of time ordered fields (I've labelled the fields from the interaction terms with $z$):

$$ \langle 0|\mathcal{T}\left\{\phi(x)\cdots\phi(x^{(b)})\phi(y)\cdots\phi(y^{(b')})\phi(z)\cdots\phi(z^{(b'')})\right\}|0 \rangle $$

Then you use a bit of technical mathematics called Wick's Theorem to show that this thing reduces to a product of two-point Green's functions (a.k.a. propagators):

$$ \langle 0|\mathcal{T}\left\{\phi\phi\right\}|0 \rangle\langle 0|\mathcal{T}\left\{\phi\phi\right\}|0 \rangle\langle 0|\mathcal{T}\left\{\phi\phi\right\}|0 \rangle\cdots $$

Now, obviously there's more than one way to pair off the $b+b'+b''$ fields you have into Green's functions, which is why you have multiple Feynman diagrams for each order in the coupling constant. Each one of these Green's functions in the product corresponds to a line in a Feynman diagram, and each way to pair the fields off corresponds to a different diagram (up to symmetry factors etc). Loops are when you get Green's functions where both fields are at the same spacetime point:

$$ \langle 0|\mathcal{T}\left\{\phi(z)\phi(z)\right\}|0 \rangle $$

because obviously you get a line from a point back to itself -- a loop. This results in unconstrained momentum because a typical Green's function has the form:

$$ \langle 0|\mathcal{T}\left\{\phi(x)\phi(y)\right\}|0 \rangle = \int \mathrm{d}^{3}p\ \frac{i}{p^2 - m^2}e^{-ip(x-y)} $$

The exponential then reduces to a delta function when you integrate over position, allowing you to easily integrate over momentum to essentially select the momentum you want (this comes from one of the integrals I left out earlier). But if $x=y$ then the exponential reduces to 1, meaning your momentum is unconstrained and generally this also leaves the integral divergent. This is the root of ultraviolet divergences in QFT, and we use regularisation to renormalise certain constants (like the mass) and absorb the infinities into unmeasureable quantities.

Edit: I notice you say you're reading P&S as well as Schwartz. I don't know Schwartz's book at all, but in P&S they derive the LSZ Reduction formula for scalar fields in Chapter 7. It's quite instructive to look through it because it gives you a much better appreciation for the form of the Feynman rules.

Even though $\hbar$ is a fundamental physical constant (and you can set it to 1 if you want), it still makes sense to talk about quantum mechanics in the limit $\hbar \to 0$. Physically, this means that the typical length scale of the system is much larger than the de Broglie wave length of all particles, so in the end what you get is classical mechanics with some additional corrections, which we call semiclassical. Considering $\hbar$ a variable, sending $\hbar \to 0$, and expanding is just a convenient mathematical way of talking about this limit.

As others have pointed out, though, Feynman diagram expansions are actually done with respect to the coupling parameter, the factor that multiplies the interaction term in the Hamiltonian. The expansion in $\hbar$ is different, and is actually the WKB approximation. You need to be careful when interpreting this sentence from Schwartz's book. Even though the expansions are different, in the case that the coupling goes to zero, we have a free theory, where the WKB approximation is exact.

Because of this, it is actually the coupling term that introduces errors in WKB, so you can also think of the calculation using Feynman diagrams at a given order as a semiclassical correction.

Physically, tree level diagrams are approximations that do not consider self-interaction. For example, if you have an electric field acting on an electron, discarding self-interaction would be writing down the Lorentz force and calculate its trajectory using the equation of motion. This is an approximation because the electron also has an electric field, and feels its effect when it is moving. Loop integrals are the way to express these self-interaction effects in perturbative QFT.