Let $A$ be a ring and let $N\subseteq M$ be two $A$-modules. If $M/N$ and $N$ are finitely generated $A$-modules, is it true that $M$ is finitely generated?
If, under the hypothesis above, $M$ is not finitely generated, assume that $S\subseteq M$ is a set of generators for $M$. Called $\overline S$ the image of $S$ in $M/N$, we have a finite $\overline T\subset \overline S$ that generates $M/N$. If, for every $\bar t\in \overline T$, we choose a $t\in M$ such that its image in $M/N$ is $\bar t$, and call $T$ the set of such $t$, it's clear that every $x\in M $ is congruent modulo $N$ to a linear combination of elements in $T$. This means that, assuming that $U$ is a finite set of generators of $N$, the set $T\cup U$ is finite and generates $M$.
Is this proof ok? I wouldn't be surprised if I missed something. Is this fact equivalent to say that, if an exact sequence $$0\to M'\to M\to M''\to 0$$ of $A$-modules, $M'$ and $M''$ are finitely generated, also $M$ is? (Exercise 2.9 of Atiyah-Macdonald)
You are right about the exact sequence. Your argument is mostly correct but a little flawed: it is not true in general that if a module is finitely generated, then any generating set has a finite generating subset (that is true for Noetherian modules, so for instance it is true over Noetherian rings). So the existence of $\overline{T}\subset \overline{S}$ in your proof is not guaranteed.
This being said, you may notice that your proof does not actually use $S$ at any point, so you may just start with any $\overline{T}$ which generates $M/N$.