How do I think of a measurable function?

I'm going to dissent from some of the comments and advise against thinking about measurability in terms of a topological property like continuity, even for heuristic purposes. The fact is that measurability is not a topological concept, and it's important to be conceptually clear about this, especially if you go on to study more abstract measure theory in which measurable spaces do not always come equipped with a natural topology (unlike the real numbers, for instance). There are, of course, important results linking measurability and continuity. I mentioned two in my comment: Lusin's theorem and Lebesgue's criterion for Riemann integrability. You should know these results and understand why they hold, but you should not conclude on their basis that, in general, the concept of measurability can be (approximately) reduced to the concept of continuity.

Now, as for intuitions, my advice is to always keep in mind that assertions about measurability are always relative to particular sigma-algebras. As you mention, a function $f: (X_1, \mathcal{F}_1) \to (X_2, \mathcal{F}_2)$ between two measurable spaces is $(X_1, \mathcal{F}_1, X_2, \mathcal{F}_2)$-measurable (notice the qualification of "measurable") provided $$f^{-1}(B_2) \in \mathcal{F}_1$$ for all $B_2 \in \mathcal{F}_2$. Sometimes mention of one or both sigma-algebras is suppressed when the context makes clear what measurable spaces one is working with. For example, in real analysis one often deals with functions from $(\mathbb{R}, \mathcal{L})$ into $(\mathbb{R}, \mathcal{B})$, where $\mathcal{L}$ is the collection of Lebesgue measurable sets and $\mathcal{B}$ is the Borel sigma-algebra.

The intuition behind the formal definition is simply that the function $f$ can be measured. That is, if we now equip $(X_1, \mathcal{F}_1)$ with a measure $\mu$, the measurability of $f$ guarantees that any "reasonable statement" $B_1$ about the values that $f$ takes is in fact a set in $\mathcal{F}_1$, and hence $\mu(B_1)$ makes sense. Returning to the canonical real analysis setup, $B_1$ might be a statement like "$f(x) \leq 5$" or "$f(x) \in (0, \infty)$". Measurability guarantees that these rough "statements" are actually sets in $\mathcal{F}_1$ that can be measured. For example, the first statement corresponds to the set $B_1 = \{x \in \mathbb{R}: f(x) \leq 5 \}$.

With this in mind, let me explain why I recommend not conflating measurability and continuity. Consider a function $f: (\mathbb{R}, \mathcal{F}_1) \to (\mathbb{R}, \mathcal{B})$, where $\mathcal{F}_1$ is the trivial sigma-algebra $(\emptyset, \Omega)$. From the formal definition, we conclude that $f$ is measurable if and only if it is a constant function (verify this). But this rules out many "well-behaved",and, in particular, continuous, functions! In other words, there are many continuous functions that are not $(\mathbb{R}, \mathcal{F}_1, \mathbb{R}, \mathcal{B})$-measurable. We see that a lot turns on the choice of $\mathcal{F}_1$.

Another important example along these lines is that there are continuous functions $f: (\mathbb{R}, \mathcal{L}) \to (\mathbb{R}, \mathcal{L})$ that are not measurable. See the discussion here. So we see that a lot turns on the choice of sigma-algebra in the codomain as well.

Finally, in response to your question about limits, it's a useful exercise to prove, directly from the definition, that the pointwise limit of a sequence of real-valued measurable functions is measurable. (As PhoemueX pointed out in the comments, pointwise convergence doesn't make sense for functions between general measurable spaces, so we now consider functions taking values in $(\mathbb{R}, \mathcal{B})$.) This sort of exercise should help your intuitions quite a bit. Consider $f_n: (X_1, \mathcal{F}_1) \to (\mathbb{R}, \mathcal{B})$, where $(X_1, \mathcal{F}_1)$ is an arbitrary measurable space, each $f_n$ is measurable, and $f_n \to f$. First, it suffices to show that $f^{-1}((-\infty, x]) \in \mathcal{F}_1$ for all $x \in \mathbb{R}$ because sets of the form $(-\infty, x]$ generate $\mathcal{B}$ (verify). Now try to show that $$f^{-1}((-\infty, x]) = \cap_{m=1}^\infty \cup_{n=1}^\infty \cap_{k=n}^\infty \{ f_k^{-1}((-\infty, x + 1/m])\}.$$ (Just think about the definition of a limit and translate the quantifiers "for all" and "for some" to $\cap$ and $\cup$, respectively.) Do you know how to conclude from here?