I have defined the boundary map $\partial$ successfully, but I am having trouble checking that it is a module homomorphism. In the argument, I used the surjectivity of a map to obtain a non-empty pre-image, but this blocks my argument that it is a module homomorphism.
Now I'm starting to wonder that perhaps it may not be a homomorphism but just a map with property $\ker\partial=\operatorname{im}g_*$ and $\ker f_*=\operatorname{im}\partial$ because my book uses the term "map" instead of "homomorphism" when stating the zig-zag lemma (so does Wikipedia).
If the diagram you started with (two short exact sequences and maps from one to the other) consists of module homomorphisms, then the zig-zag map is also a module homomorphism. The proof has two ingredients.
First, you need that the zig-zag is a well-defined function. That is, the result depends only on the input, not on your choice of a pre-image with respect to that surjection you mentioned. This is shown by considering what happens if you make two different choices and follow the difference between them through all the subsequent steps in the zig-zag.
Once you have this well-definedness established, the second step, showing that the function is a homomorphism, is rather easy. Consider, for example, what happens when you apply the zig-zag construction to two elements $x$ and $y$ and their sum $x+y$. You choose any pre-images, say $x'$ and $y'$, for $x$ and for $y$ with respect to that surjection, but then, instead of choosing a pre-image for $x+y$ arbitrarily, you choose $x'+y'$. (You can safely do that because of the first part of the argument --- any choice would get you the same final result, so you can make the choice to simplify the argument.) Then the rest of the zig-zag preserves the fact that what you're getting from $x+y$ is the sum of what you got from $x$ and what you got from $y$. In particular, the final results respect addition. A similar (but slightly easier) argument shows that they also respect multiplication by elements of the ring of scalars. So you have a homomorphism of modules.