Let's illustrate with an example. Say $X$ is the blow-up of $\mathbb P^2$ at a point, with exceptional divisor $E$. Think about the line bundle $\mathcal O_X(H+E)$, where $H$ is a line not through the blown-up point. I think it's easier to think about this in the language of divisors and linear systems, but you can translate it back to line bundles.
What does the linear system $|H+E|$ look like in our case? You can check that a divisor in this linear equivalence class consists of $E$ together with some line $L$ (or $2E+R$, where $R$ is the strict transform of a line through the blown-up point). In particular, $E$ is in the base locus, and the fixed part $F$ is just $E$.
What's $dim H^0(X,\mathcal O_X(F))$? There's no way it can be greater than 1 -- if it were, it would mean that $F$ moves in a linear system of dimension at least $1$. But that means that at least one component of $F$ isn't in the base locus of $F$ at all (it's not fixed), and there's no way such a component could be in the base locus of $L$. So it shouldn't have been included in $F$ in the first place.
At the level of linear series, the map $L(-F) \to L$ just says to take an element of $L(-F)$, add on $F$, and you get something in the linear series $|L|$. In our case, something $|L(-E)|$ will just be a line, and you get something in $L$ by throwing in $E$ too.
Your interpretation at the end seems correct to me. Note that there's always a map $H^0(X,L(-F)) \to H^0(X,L)$ whenever $F$ is effective, but it's only an isomorphism if $F$ is the fixed part.