Gaining intuition for how submodules behave I'm studying elementary commutative algebra this semester, largely following Atiyah-MacDonald. I often find myself in a situation where I'm interested in whether some property of an R-module M is inherited by its submodules (e.g. the property of being finitely generated over R) and I feel like I am lacking the necessary intuition to decide whether something is the case or not. In the case of finite-generation, for example, I think that my mental picture of a finitely-generated R-module is still too close to that of a finite-dimensional vector space to be able to intuit the fact that this simply shouldn't hold true in general. 
So here's my question - when you run across some property for a module and you want to know whether this is inherited by its submodules, how do you begin to think about the problem? Do you have a standard stock of counterexamples (or a procedure of sorts to concoct a counterexample)? Or do you have a more nuanced way of informally thinking about modules that captures more of their behavior? As I only have a semester of commutative algebra under my belt (think Atiyah-MacDonald), I'd appreciate answers that tend towards the more elementary end of the subject, although if you think that it's not possible to gain a good feel for the way that modules behave without diving deeper, I'd like to hear that too. 
 A: My not-so-expert opinion would be that "What properties of M does a submodule of M inherit" may not be a general enough question to have a helpful answer. There are a few such properties that come up frequently (especially the Noetherian property that you mention), and it's worth understanding each on its own. (For example, proposition 6.3.1 on page 75 of A-M is vital.) I agree with you that thinking up examples of modules, to give yourself some intuition, is useful.
To that end, I'd suggest taking a look at the question "What representative examples of modules should I keep in mind?" Unfortunately, many of the answers to that question may be a little too pathological for your taste, and the accepted answer (to which BCnrd also refers) requires the motivation of algebraic geometry.
I like the suggestion in Andrew Critch's answer to "find interesting rings." A module over $R$ is often a tool to study $R$, rather than a fundamental object in its own right. Many "interesting" rings might seem perfectly natural objects of study, even to a beginning student: for example, a polynomial ring, or a polynomial ring modulo an ideal, $k[x_1,\ldots,x_n]/I$.
To get comfortable working with these rings (and modules over them), I'd suggest learning some algebraic geometry; that way you can think of a ring like $\mathbb{C}[x, y]/(y^2-x^3)$ as the curve $y^2=x^3$ in the plane, rather than some kind of fancy made-up thing. Meanwhile, I agree with BCnrd that doing the exercises in A-M (or taking an algebra course and doing the assignments) is a good way to start getting more familiar.
ETA: Failed to mention that the most important examples of modules over a ring $R$ are ideals in $R$. Luckily, Hailong Dao's excellent answer has this covered, so rather than remark further, I'll just say, read that.
A: 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 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 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 quite far). I hope you will have some fun! 
