Let $X_0$ be a connected scheme defined over $\mathbb F_{p}$ and let $X$ be the product $X_0 \times_{\mathbb {F}_p} \overline{\mathbb F_p}$, as usual, with the natural map $\pi:X \to X_0$. Given an etale sheaf $\mathcal G_0$ on $X_0$, let $\mathcal G$ be the pullback $\pi^* \mathcal G_0$. The Frobenius $F: X \to X$ induces, by adjunction induces a natural map $\mathcal G \to F_* F^* \mathcal G$ which in turn induces $H^i(X, \mathcal G) \to H^i(X, F^* \mathcal G)$. The key ingredient to getting the Frobenius action on the cohomology is to get a map $F^*\mathcal G \to \mathcal G$, which is an isomorphism. It is this map I have trouble understanding. In Bhargav Bhatt's seminar here, at the end of page 8 , we see the desired map being defined roughly as follows:
Firstly there are the two projection maps $\text{pr}_1$ and $\text{pr}_2$, $X\times_{X_0}X \to X$, and they satisfy $\pi \circ \text{pr}_1=\pi \circ \text{pr}_2$. From this, we get $\text{pr}_1^* \circ \pi^*\mathcal G_0=\text{pr}_2^* \circ \pi^*\mathcal G_0$, which means that $\text{pr}_1^* \mathcal G$ and $\text{pr}_2^*\mathcal G$ are equal on the nose. However, there it is claimed only that $\text{pr}_1^* \mathcal G \cong \text{pr}_2^* \mathcal G$.
Next, $X\times_{X_0}X$ is claimed to be $\text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)\times X$. I don't see why this should be. A typical element of $X\times_{X_0}X$ is a pair $(x,y)$ such that $\pi(x)=\pi(y)$. But since $X_0$ is the set of Galois orbits of $X$, this means that there is an automorphism $\sigma \in \text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)$ such that $y= \sigma(x)$. But this $\sigma$ is defined only up to $\text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_{p^{\text{deg}(x)}})$, so I don't see why the product structure should hold.
Even if I assume the above two assertions, how does the isomorphism $\text{pr}_1^* \mathcal G \cong \text{pr}_2^* \mathcal G$ decompose into $\text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)$-many distinct isomorphisms?