I think of them pretty simply as differential forms with zeros in the denominator of the "coefficient function" upon choosing a local uniformizing parameter (which is really just the definition).
Edit: The following isn't quite right, as pointed out in the comments. See below for attempt at fixing it.
_____
However, if you want something more akin to your "map to $\mathbb{P}^1$" description, you might try something like this: to the invertible sheaf $\Omega$ you can associate a projective bundle $\mathbb{P}(\Omega)$ equipped with a map $\pi:\mathbb{P}(\Omega)\to C$ whose fibers are the projectivizations of the fibers of $\Omega$. Then a meromorphic differential form should correspond to a section $s$ to $\pi$ and the poles of the form are the points where $s(x)=\infty$.
I can't say that I've ever seen exactly this written down, but it seems quite reasonable...
_____
Perhaps one should consider the projectivization of the bundle $\Omega\oplus\mathcal{O}$ instead. Clearly if I take a regular differential form $\omega$, it gives rise to a section of $\mathbb{P}(\Omega\oplus \mathcal{O})$ by considering the image of $\omega\oplus 1$ in the projectivization.
Arguing locally, it seems that if I take a meromorphic differential form of the form $t^{-n}udt$ where $u$ is a local unit, I can associate to it the image of $udt\oplus t^n$ in the projectivization. Now glue over the curve to associate a section of this projective bundle to your chosen meromorphic differential form. If everything glues without incident, it seems that the resulting section should have the desired property that the poles are the points mapping to $\infty$ in the fiber, just by construction.
Hopefully this makes more sense than my first attempt :)
