In section 19.9.5 of Vakil, he says an elliptic curve $E$ with a point $P$ admits a double cover of $\mathbb{P}^1$ given by the line bundle $\mathscr{O}_E(2P)$. Then he says $P$ is a branch point because one of the sections of $\mathscr{O}_E(2P)$ vanishes at $P$ to order 2, so there is a point of $\mathbb{P}^1$ whose preimage consists of $P$ with multiplicity 2.
Here's my understanding: let $s$ be a global section of $\mathscr{O}(2P)$ that only has a zero of order 2 at $P$, and let $t$ be a global section of $\mathscr{O}_{\mathbb{P}^1}(1)$ that pulls back to $s$. Then the locus of where $s$ vanishes should be the pullback of the locus of where $t$ vanishes, which means there must be a point in $\mathbb{P}^1$ whose preimage is just $P$. Is that right?
But the definition of the global sections of $\mathscr{O}(2P)$ is $$\{t \in K(E)^{\times}: div \, t + 2P \geq 0\} \cup \{0\}$$ which means rational functions that can only have poles at $P$ of order less than equal to 2. Then can't we have a global section that only has a zero of order 2 at $Q$ for any point $Q$?
The confusion might stem from the fact that, taking the zero locus of a global section $s\in H^0(X,\mathscr{O}(D))$ is different from taking the divisor of zeros of the corresponding rational function.
As an example, consider $X=\mathbb{P}^1$ and $\mathscr{O}(p)$ for some point $p$. This is isomorphic to $\mathscr{O}(1)$, which has global sections degree 1 homogeneous polynomials. In particular, they vanish at a single point (except for the zero section). On the other hand, you can cook up rational functions $t$ such that $div t + p \ge 0$, where $t$ has a lot of zeros!
Anyway, one way to resolve your problem is to consider the Cartier divisor associated to $2P$ where you can consider $U_1 = E-P$ with a function $1$, $U_2$ to be a open containing $P$ with a function defined on $U_2$ which vanishes at $P$ to order $2$. Now if you consider the global section $1$, then it vanishes at $P$ of order $2$.