If $M$ is finitely generated then $M/N$ is finitely generated.

Suppose $M$ is generated by $n$ elements, say $\{a_i\}_{i=1}^n$.

Claim : $M/N = (a_i + N \ |\ i \in [1,n])$.

$(\supseteq)$ It is clear.

$(\subseteq)$ Suppose $x + N \in M/N$ with $x = \sum r_i a_i$. Then $x + N = \sum r_i (a_i + N)$, as desired. Thus $\{a_i + N\}_{i=1}^n$ is a generating set for the $R$-module $M/N$.