The rank of an $R$-module is the maximum number of $R$-linearly independent elements in the module.
If $R$ (commutative with identity) is an integral domain and $M$ is a finitely generated $R$-module then $\operatorname{rank}(M)=\operatorname{rank}(N)+\operatorname{rank}(M/N)$, where $N$ is an $R$-submodule of $M$.
If it was given $M/N$ is a free $R$-module then we can write $M$ as a direct sum of $N$ and $M/N$. Here we can only say that $M/N$ is finitely generated as $M$ is finitely generated. How do I go further?
Just an elaboration. Let $R$ be a domain and $M$ any module. Then rank of $M$ is equal to the dimension of the vector space over $K$, the fraction field of $R$ of $M\otimes_R K$ (if you are unfamiliar with tensor products, you can also use localization). So, rank of $M$ is rank of $M\otimes K$ which is clearly the sum of the ranks of $N\otimes K$ and $M/N\otimes K$ and the rest should be clear.