Question: Let $R$ be a commutative ring with $1\neq 0$. What do the $R$-submodules of $R^n$ look like?

Answer: 
Suppose $R$ is an integral domain. Then $R^n$ is a torsionfree $R$-module, hence all of its submodules are torsionfree.

If $M$ is a finitely generated torsionfree $R$-module, then $M$ is isomorphic to a submodule of some $R^n$: Lemma 21.8 [of these notes](http://alpha.math.uga.edu/%7Epete/integral.pdf). (The proof is easy: since $M$ is torsionfree it embeds in $M \otimes\_R K \cong K^n$, where $K$ is the fraction field of $R$. Since $M$ is finitely generated we can clear denominators.)

If $R$ is Noetherian then every submodule of $R^n$ for some $n$ is finitely generated, so the above gives a characterization of the submodules of finitely generated free modules. If $R$ is not Noetherian, then $R$ itself has some infinitely generated submodules.
In general, you can't say much. 

The only thing I can think of to say is that for commutative principal ideal domains, you can say that the submodules are also free. (Notice this also contains the "obvious" case of fields.)

I know that the submodules also have to be of lesser or equal rank in the finite rank case. I'm not sure if it's true in the infinite rank case, but I suspect so.