Let $R$ be a ring and $I \subset R $ a two-sided ideal and $M$ an $R$-module with
$$IM = \{ \sum r_i x_i \mid r_i \in I, x_i \in M \}. $$
Show that if the $R$-module $M$ is free of rank $n$ then $M/IM$ is also an $R/I$ free module of rank $n$. I have shown that $IM$ is an $R$-submodule of $M$ and an $R/I$ module but i do not know how to approach this question. Any assistance is greatly appreciated.
Your proof that $M/IM$ is a submodule of $M$ is surely wrong. Take $R=M=\mathbb{Z}$ and $I=2\mathbb{Z}$. Then $M/IM=\mathbb{Z}/2\mathbb{Z}$ that doesn't even embed in $M$.
Saying that $M$ is a free module of rank $n$ is the same as saying that $M\cong R^n$ and it's not restrictive to take $M=R^n$. Now $IR^n=I^n$ and $$ M/IM=R^n/I^n\cong(R/I)^n $$ as $R$-modules via the map $$ (x_1,x_2,\dots,x_n)+I^n\mapsto(x_1+I,x_2+I,\dots,x_n+I) $$ This is also an isomorphism of $R/I$-modules, as you can verify.