Let $G = G_1 \ast G_2$. Then $$G/[G,G] \simeq G_1/[G_1, G_1] \oplus G_2/[G_2,G_2].$$
This is Munkres, section 69, exercise 1; with the hint to use the extension condition for free groups and direct sums in each direction, so find 2 homomorphisms, and show that they are mutual inverses.
It is easy to find some homomorphism in either direction:
(i) left-hand-side (LHS) to right-hand-side (RHS): as RHS is the direct sum of two abelian groups, it is abelian. Any homomorphism from $G$ to RHS will then induce a homomorphism from LHS to RHS (eg, from the elementary first isomorphism theorem).
(ii) RHS to LSH: eg, mapping $x[G_1, G_1]$ to $x[G,G]$ is a homomorphism that extends to one on $G$, using the extension condition for direct sums.
However, (i) is just an abstract statement from which I do not see how to make progress in showing that we have mutual inverses (nor does (ii) help me much so far).
Note: at this point in the book, Munkres has only introducted fundamental groups, covering spaces, deformation retracts, etc, plus a few chapters on free groups and direct sums, to the level of showing the fundamental groups of the Torus, some discussion on figure 8 etc. There are probably very abstract ways in which this might follow; but if you could kindly stick to as elementary a proof or hint as possible, I would appreciate! My group theory knowledge is currently at about the level of one class of group theory.
P.S.: This is not homework. I'm reading Munkres for fun, and try to do every exercise. Usually, this goes fine. But I'm really stuck here.
I only managed to find a direct proof. Let $$N:= [G, G], \quad N_i = [G_i, G_i].$$ The key fact to use is that $G/N$ and $G_i/N_i$ are abelian. Call this (1).
Note first that in the finite case for abelian groups $G, H$, we have $G \bigoplus H = G \times H$. This is the notation we will use. Let
$$\begin{align} \phi: & G/N \rightarrow G_1/N_1 \times G_2/N_2 \\ &xyN \mapsto (xN_1, yN_2) \\ \end{align}$$ $$(\phi(\prod x_iy_i N) = (\prod x_i N_1, \prod y_i N_2)),$$
which, using (*), is well-defined, and an obvious surjective homomorphism.
To see that $\phi$ is injective, note first that $$e_{G/N} = N, e_{G_i/N_i} = N_i, e_{G_1/N_1 \times G_2/N_2} = N_1 \times N_2.$$
Assume $x y \tilde{x} \tilde{y} N \mapsto N_1 \times N_2$. Then $x\tilde{x} \in N_1 \subset N, y\tilde{y} \in N_2 \subset N$, so, using (1), $$xy\tilde{x}\tilde{y}N = x\tilde{x}y\tilde{y}N= N.$$
Similarly, if $\prod_{i=1}^nx_iy_iN \mapsto N_1 \times N_2$, we have (using (1)) $$\prod_{i=1}^{n}x_i \in N_1 \subset N, \quad \prod_{i=1}^{n}y_i \in N_2 \subset N,$$ so, using (1) again, $$\prod_{i=1}^{n} x_iy_iN = (\prod_{i=1}^{n}x_i)(\prod_{i=1}^{n}y_i)N = N.$$
We conclude that $\phi$ is injective as well, and so defines the desired isomorphism.