Let $X$ be an abelian surface and $L$ be a very ample line bundle on $X$. Let $M\in \textrm{Pic}^o\,X$. That is $T_x^*M\simeq M$ for all $x\in X$. We know that $H^i(X,M)=0$ for all $i$.
Is it true that $h^0(X,L) = h^0(X,L\otimes M)$? How do we prove this?
The fact that you are looking for is true. It comes from two facts.
First, the Euler characteristic is a numerical invariant. Thus, if you twist $L$ by a numerically trivial line bundle $M$, you get the same Euler characteristic. You can see this form Riemann-Roch for surfaces: since $M$ is numerically trivial, intersection product with $L$ or $L\otimes M$ are the same.
Also, if you twist an ample line bundle $L$ by a numerically trivial line bundle $M$, $L \otimes M$ is still ample. See Kleiman's criterion for ampleness about this.
Second, for an ample line bundle $H$ on an abelian variety, we have $\chi (H)=h^0(H)$. This follows from a more general result about abelian varieties, which says that if a line bundle $L$ is non-degenerate (i.e. $\chi(L) \neq 0$), then all the cohomologies but one vanish. As for reference, see for instance Mumford's book "Abelian Varieties", under the section "Cohomology of Line Bundles".
These two facts give you the claim. Also, note that we just need $L$ ample, not very ample.