So let $(E,p)$ be an elliptic curve over a field $k$ with a choice of $k$-valued point $p$. Then by Riemann-Roch, there are two global sections of $\mathcal{O}_{E}(2p)$ which gives a double cover of $\mathbb{P}^{1}$ and I let this morphism be called $\phi$. Then one of the sections which give the map to $\mathbb{P}^{1}$ vanishes at $p$ of order $2$, meaning that the preimage of $0$ is one point, so $p$ is a branch point. Write this section as $\sigma_{p}$.
Then by Riemann-Hurwitz, there are three other branch points, so we pick $q$ to be one of them.
Why do we have that $2p\sim 2q$ as divisors (reference [Hartshorne IV Example 4.8.2], [Vakil, Ex 19.9.A])?
The only way I can think of is that by the same argument, $\mathcal{O}_{E}(2q)$ also has 2 sections, one of which vanishes at $q$ of order 2. Write this as $\sigma_{q}$.
Now, I am very tempted to say that hence $2p-2q$ is principal because we can take $\sigma_{p}/\sigma_{q}$. But I think this proof is missing some crucial steps because what if I take a non-branched point say $z$ of the morphism $\phi$ and run the whole argument, I will arrive at the same conclusion... which looks to me absurd...
What should be the correct argument in this case?
A more elaborate explanation of my doubt
I should be more precise with my question: I was thinking of what happens if instead we take $\mathcal{O}_{E}(2z)$. I am afraid of a contradiction here because we already know that $z$ is not a branched point of $E$, and as you have said any sections $\tau$ that vanishes at $z$ must vanish at $z'$ also. Taking into account that $\deg div(\tau)=2$, you can't have $\tau$ that vanishes at $z$ of order 2.
On the other hand, if we choose $(E,z)$ as a elliptic curve instead of $p$, by Riemann-Roch, we have $\mathcal{O}_{E}(2z)$ having two global sections? one vanishing at $z$ at two points again. Are we talking about two different elliptic curves, or two different coverings $E\rightarrow\mathbb{P}^1$?
If you take a non branched point $z$, you can find a global section that vanishes simply at $z$ but will also vanish at another point $z' \neq z$ (because $\phi^{-1}(\phi(z)) = \{z,z'\}$). Then $2p \sim z + z'$. There is no absurdity there.