What are the submodules of $R^n$? Let $R$ be a commutative ring with $1\neq 0$. What do the $R$-submodules of $R^n$ look like?
 A: I don't think there is any nice characterization.  The submodules of $R^1$ are just ideals of $R$.  There are then a nice class of submodules of $R^n$ of the form $I_1 \times I_2 \times \cdots \times I_n$ where $I_j$ are ideals of $R$.  However, these do not exhaust the submodules.  For example, the subgroup of $\mathbb Z \times \mathbb Z$ generated by $(1,1)$ is not of this form.
If you are just getting used to working with rings and modules, I think the question to ask is  not "What do submodules of $R^n$ look like," which is a little too broad (although there is no way to know that without having experience), but rather "What are some examples of $R$-modules or of submodules of $R^n$ which contradict what my intuition might be if I am used to working with abelian groups and vector spaces."
Unfortunately, I can't think of any good examples which are weird but not too weird right now.  However, it might be worthwhile to try to answer your question when $R=\mathbb C[x]/x^2$, which is in some sense the simplest example of a ring which is neither a field nor a quotient of $\mathbb Z$.  $R^2$ is $4$-dimensional as a $\mathbb C$-vector space.  Can you find all the submodules (which will necessarily be vector subspaces)?  There is a vector space basis of $(1,0), (0,1), (x,0), (0,x)$.  Is there a nice characterization of which vector subspaces are submodules?
A: 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.  (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.
A: 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.
