I am currently dealing with this result from William Stein's Algebraic Number Theory notes, and I also underlined the part of which I am not sure yet:
I know that $g(M_0)$ is trivial since we have a short exact sequence and $M_0$ lies in the image of $f$. So it is $g(M'/M_0) \simeq g(M')$. The latter is a submodule of $N$. But how can I make the transition to $M'/M_0$ (I only showed that its image is isomorphic to some submodule of $N$)?
I think it is probably just some small detail but I am currently at loss.
Restrict $g$ to a map $g':M'\to N$. The kernel of $g$ is $M_0$ so by the First Isomorphism Theorem $g'$ sets up an isomorphism $M'/M_0\to g'(M')=g(M')$, a submodule of $N$.