Let $X$ be a smooth projective complex surface, $L$ a line bundle decomposed in its fixed and moving part as $|L|=F+|M|$.
Intuitively, the inclusion of $|M|$ into $|L|$ yields an isomorphism $H^0(X,M)\simeq H^0(X,L)$.
(but I am not sure how to show this precisely, so any elaboration would be much appreciated)
What happens for higher cohomology (possibly adding some assumptions on $L$) ?
For example, do we have $h^1(M)=h^1(L)$ ? (if not, what if $h^1(L)=0$?)
Firstly, the isomorphism you state is more or less the definition of the fixed part of $L$. More precisely, I guess one could say that the fixed part of $L$ is the maximal divisor $F$ such that $H^0(X,F)$ is 1-dimensional and the multiplication map
$$H^0(X,L-F) \otimes H^0(X,F) \rightarrow H^0(X,L)$$ is an isomorphism.
To see what happens with higher cohomology, look at the ideal sheaf sequence for the divisor $F$:
$$\mathcal O_x(-F) \rightarrow \mathcal O_X \rightarrow \mathcal O_F.$$
Tensor by $L$ and take cohomology:
$$ 0 \rightarrow H^0 (X,L-F) \rightarrow H^0(X,L) \rightarrow H^0(F,L_{|F}) \\ \rightarrow H^1(X,L-F) \rightarrow H^1(X,L) \rightarrow H^1(F,L_{|F}) \\ \rightarrow H^2(X,L-F) \rightarrow H^2(X,L) \rightarrow 0.$$
(where the final $0$ is $H^2(F,L_{|F})$, which vanishes because $F$ is 1-dimensional.)
This is pretty messy, so let's say we're on a surface with Kodaira dimension at most 0. Then Serre duality shows that the $H^2$ terms in the above sequence vanish. That simplifies things to some extent.
Now by definition of "fixed part", the first nontrivial map in the above sequence is an isomorphism, so we reduce to a 4-term exact sequence
$$ 0 \rightarrow H^0(F,L_{|F}) \rightarrow H^1(X,L-F) \rightarrow H^1(X,L) \rightarrow H^1(F,L_{|F}) \rightarrow 0.$$
Now we can see that various things can happen, but we only get the isomorphism you want if $L_{|F}$ has no cohomology.
