Suppose that $X$ is a curve and $f\in K(X)$. Then $f$ can be realized as a map $f:X\to\mathbb{P}^1$ given by $x\mapsto[f(x):1]$. My question is: if we take any point $a\in\mathbb{P}^1$, are the points in the fiber $f^{-1}(a)$ linearly equivalent as divisors?
Suppose $P_1,P_2$ are two divisors in $f^{-1}(a)$. Then I want that $P_1-P_2=\sum v_p(f)P$, but I'm not sure how to get this from the fact that $P_1,P_2$ both map to $a$. What am I missing?