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?
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.