Every line bundle $L$ on a complex algebraic curve $X$ is of the form $\mathcal{O}(D)$, where $D$ is some divisor on $X$. This means $L$ has at least one nonzero meromorphic global section, i.e. $$H^0(L \otimes_{\mathcal{O}_X} \mathcal{M}_X) \neq 0$$ where $\mathcal{M}_X$ is the sheaf of meromorphic functions on $X$.
Is there any way to see this by directly calculating global sections of the sheaf $L \otimes_{\mathcal{O}_X} \mathcal{M}_X$ and without assuming $L \cong \mathcal{O}(D)$ for some divisor $D$? I've seen other arguments so I'm mostly just interested in a calculation of the above cohomology group.
More generally, how does one compute $H^0(V \otimes_{\mathcal{O}_X} \mathcal{M}_X)$, where $V$ is an arbitrary vector bundle on $X$?
Here's one way to show that $L$ has a global meromorphic section: show that $L\cong\mathscr O_X(D)$ for some divisor $D$. Let $L(D)=L\otimes_{\mathscr O_X}\mathscr O_X(D)$ for any divisor $D$. Mimicking the proof of Riemann-Roch, one can show that $$h^0(L(D))-h^1(L(D)) = h^0(L) - h^1(L) +\deg D.$$ In particular, fixing some point $p\in X$, $h^0(L(kp))\ge h^0(L(kp))-h^1(L(kp))=1$ when $k=h^1(L)-h^0(L)+1$. Taking a nonzero holomorphic section $s\in H^0(L(kp))$ and letting $D=(s)$, we see that $L(kp)\cong\mathscr O_X(D)$, so $L\cong\mathscr O_X(D-kp)$, and now we see that it must have some global meromorphic section.