Here is the definition from Hartshorne:
If $f:X \to Y$ is a finite morphism of nonsingular curves. For any $Q \in Y$, let $t \in \mathcal{O}_{Q}$ be a local parameter at $Q$, i.e., $t$ is an element of $K(Y)$ with $v_{Q}(t)=1$, where $v_{Q}$ is the valuation corresponding to the DVR $\mathcal{O}_{Q}$. Define $f^{*}Q = \sum_{f(P)=Q}v_{P}(t)P$.
Now in practice, if I have a rational function $f$ on a curve $X$, which has pole at $P$ of order $n$, I can take $f$ as a morphism $X \to \mathbb{P}^{1}$ sending $P$ to $\infty$. Then I want to compute $f^{*}P$. In the above definition I have to find that local parameter $t$ and the valuation under $v_{P}$, but I don't have any idea (but I believe the pullback is $nP$). How to do it explicitly?
Also, is there any alternative definition of pullback divisor which can be done easily?
Thanks for helping!
Note that in the definition of the pull-back divisor, $f^*Q=\sum_{f(P)=Q}v_P(t)P$, the integer $v_P(t)$ should be understood as $v_P(f^\sharp t)$. More precisely, there is an extension of fraction fields $f^\sharp:K(Y)\rightarrow K(X)$, and $v_P(t)$ means the valuation at $P$ of the image in $K(X)$ of $t$ by this extension.
Let $K(\mathbb{P}^1)=K(z)$ with $z$ an indeterminate. Then a local parameter at $\infty$ is $1/z$.
Let $f\in K(X)$, this induces a morphism $X\rightarrow\mathbb{P}^1$ such that the extension of fraction fields is $f^\sharp:K(z)\rightarrow K(X)$ with $z\mapsto f$. So the image of our parameter at $\infty$ by this extension $f^\sharp(1/z)=1/f$.
Now if $f$ has a pole of order $n$ at $P$, this means that $v_P(f)=-n$. Thus $$v_P(1/z)=v_P(f^\sharp (1/z))=v_P(1/f)=n$$
And finally, if $P$ is the only pole of $f$, the sum contains only one term, so $f^*\infty=nP$