Let $f:X\longrightarrow S$ be any morphism of $\mathbb{F}_{p}$-schemes. Why is the relative Frobenius $F=F_{X/S}$ flat, as claimed here, and why does this imply $H^{\bullet}(\mathbf{Pullb}(F_{S},f),F_{*}M)=H^{\bullet}(X,M)$ for any quasicoherent $\mathcal{O}_{X}$-module $M$?
This is totally against my understanding of the situation. What if, for instance, $f$ is Noetherian but not locally regular, then by Kunz's theorem $F$ would not be flat.