If M+N and M$\cap$N are finitely generated modules, so are M and N.

I assume that $M,N$ are submodules of a given $R$-module. Since $M+N$ is finitely generated, the same is true for $(M+N)/N \cong M/(M \cap N)$. Since $M \cap N$ is finitely generated, this implies that $M$ is finitely generated. By symmetry, also $N$ is finitely generated.

More explicitly (especially when you don't know quotients and isomorphism theorems yet): Let $\{m_i+n_i\}$ be a generating set of $M+N$. Let $\{u_j\}$ be a generating set of $M \cap N$. Then I claim that $\{m_i\} \cup \{u_j\}$ is a generating set of $M$ (in particular, if $M+N$ and $M \cap N$ are finitely generated, then the same is true for $M$). In fact, let $m \in M$. Then we can find $a_i \in R$ with $m = \sum_i a_i (m_i + n_i)$. It follows that $m - \sum_i a_i m_i \in M \cap N$, hence there are $b_j \in R$ with $m - \sum_i a_i m_i = \sum_j b_j u_j$, i.e. $m = \sum_i a_i m_i + \sum_j b_j u_j$, QED.

If think the exact sequence $0 \to M \cap N \to M \oplus N \to M + N \to 0$ should help to find a finite number of generators for $M \oplus N$ and thus for $M$ and $N$.