Question. For $f(x) \in \mathbb{F}_p[x]$ a monic separable polynomial, define $U=\mathrm{Spec} \mathbb{F}_p[x,y]/(y^2-f(x))$. Consider the degree 2 morphism $f:U\to \mathbb{A}^1$ given by $(x,y)\mapsto x$. Let $C$ be a nonsingular model of $U$ (i.e. a connected nonsingular projective curve containing $U$ as an open dense subscheme) defined over $\mathbb{F}_p$, and extend $f$ to a degree 2 morphism $\pi:C\to \mathbb{P}^1$. Suppose that the ramification index at $\infty\in \mathbb{P}^1$ is $2$ (which is the case if and only if $\deg(f)$ is even by Hurwitz formula, cf. the linked document in ''Background'' below), so the fiber over $\infty$ is a singleton $\{x\}$. How can one deduce that $x\in C(\mathbb{F}_p)$ and thus $\#C(\mathbb{F_p})-\#U(\mathbb{F_p})=1$? Why is it impossible for $k(x)$ to be a degree 2 extension of $\mathbb{F_p}$?
Any hint will be appreciated. Thanks in advance!
Background. This is one step in the proof of THÉORÈME 1.4.1.1. in ''Sommes exponentielles'' by Katz and Laumon, which states that, for any monic separable polynomial $f(x)\in \mathbb{Z}[x]$ whose reduction mod $p$ (for an odd prime $p$) is again separable, one has the following bound
$$ \Big|\sum_{x\in\mathbb{F}_p}(\dfrac{f(x)}{p})\mbox{ }\Big| \leq (\deg(f)-1)\sqrt{p} $$ where the summands on the left are Legendre symbols. Then the desired statement in my question appears at the bottom of page 21 in the linked document.
We have $\pi\colon C(\overline{\mathbb F}_p)\to \mathbb P^1(\overline{\mathbb F}_p)$, which is $\Gamma=\mathrm{Gal}(\overline{\mathbb F}_p/\mathbb F_p)$-equivariant, and $C(\mathbb F_p)=C(\overline{\mathbb F}_p)^\Gamma$ and $\mathbb P^1(\mathbb F_p)=\mathbb P^1(\overline{\mathbb F}_p)^\Gamma$. Thus, it suffices to check that $\Gamma$ fixes $x$. But this is clear, since $\Gamma$ fixes $\pi^{-1}(\infty)=\{x\}$.