We consider $f\colon X \to \mathbb{P}^1$ a non degenerate morphism of a smooth surface $X$ with irreducible fibers. I want to understand why $f^*: H^0(O_{\mathbb{P}^1}(1))\to H^0(O_X(F))$ is an isomorphism, where $F=f^{-1}p$.
$f^*$ is injective: We consider a section $ax+by\in H^0(O_{\mathbb{P}^1}(1))$, where $|x:y|$ are the homogeneous coordinates of $\mathbb{P}^1$. Then the pullback section will be $af_0+bf_1$, where $f_0, f_1$ are the components of $f$. Now if $af_0+bf_1=0$, then the polynomial $ax+by$ it would have an infinite number of roots, because of $f$ is non degenerate. So $a=b=0$, that means $f^*$ is injective.
Regarding $f^*$ surjective, we can say that if $C\in H^0(O_X(F))$ and if we choose $p\in C$, then we can consider the fiber $F_p=f^{-1}(f(p))$. Now $C\equiv F\equiv F_p$ and $C.F_p=C.F_q=0$, where $q\not \in C$. This means $F_p\leq C$, and so $C=F_p$.
Is it correct?