Doing all exercises in Atiyah-MacDonald, like BCnrd suggested, is surely the ideal way to learn about this and much more. Let me offer a couple of practical tips to get you started:
A surprisingly effective example to keep in mind when you deal with any question about submodules of a module $M$ is to take $M=R$. Then the submodules of $R$ are just the ideals of $R$, which are concrete enough to check your intuition, but still have possess a very rich structure so that not much is lost.
Also, since many properties of modules fail to pass to submodules in higher dimension, it usually suffices to consider some small example, say $R= k[x,y]$.
As an example, let says you are trying to understand the following question: Over what Noetherian ring $R$ is a submodule of any free module free? (this is of course true for v.spaces)vector spaces).
If you take $M=R$, it follows that all ideals $I$ have to be free. If $R=k[x]$, this is true, and already an interesting exercise, but if $R=k[x,y]$, just take $I=(x,y)$. $I$ is not free because the generators have a non-zero relation: $xy-yx=0$. This example also suggests that all ideals in $R$ have to be principal, otherwise similar counter-examples can be found. So you naturally gets to principal ideal rings.
If you want to play with it a bit more, since $R/I$ fits into an exact sequence:
$$0 \to I \to R \to R/I \to 0 $$
This says that $R/I$ has projective dimension at most $1$ for any ideal $I$. This leads you to some serious restriction on $R$, which will point you to the right condition, from a different perspective.
You can replace "free" by "locally free" and play the same game, it will naturally leads you to all sort of interesting things worth learning about commutative rings, for examples projective modules or Quillen-Suslin theorem, etc.
(There are, of course, other ways to approach this particular question, my point is by considering $M=R$ you can already get very quite far). I hope you will have some fun!
