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

Answer: 
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?
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.