Let $X$ be a smooth projective curve over a characteristic $p$ scheme $S$.
Let $\mathcal E$ be a vector bundle on $X$.
Let $\Psi:\operatorname{Der}_S(\mathcal O_X,\mathcal O_X) \to \operatorname{End}_S (\mathcal E) $ be the $p$-curvature (which is $p$-linear, i.e. additive and $\Psi(gD)=g^p\Psi(D)$) for $g\in \mathcal O_X$ and $D\in \operatorname{Der}_S(\mathcal O_X,\mathcal O_X) $.
I am reading a paper that says that the above $p$-linearity implies that
$\Psi$ defines an $\mathcal O_X$-linear morphism $\Psi':\operatorname{Der}_S(\mathcal O_X,\mathcal O_X) \to f_{X*}\operatorname{End}_S (\mathcal E) $ where $f_{X}$ is the absolute Frobenius on $X$.
I don't understand this implication, I have some intuition of it since the absolute Frobenius sends sections to their $p$-power but I can't build a rigorous thought about the role played by the direct image of the absolute Frobenius.
I appreciate your help.
The absolute Frobenius send an affine open subset to itself (actually it is constructed by glueing the absolute Frobenius on an affine cover). It follows that everything is local and we can reduce the problem to the case of an affine scheme $\operatorname{Spec}A$. The map $\Psi$ is then a map between $A$-modules such that $\Psi(gD)=g^p\Psi(D)$ for any $g\in A$.
Now the absolute Frobenius is a ring-homomorphism $f:A\to A$ such that $f(g)=g^p$. It gives $A$ the structure of an $A$-algebra and any $A$-module can be seen as a $A$-module through this map (by restriction of scalar). The structure is then the following $g.D=f(g)D=g^pD$ for any $g\in A$ and $D$ in the module.
But this is what the direct image does : on affine subset $\operatorname{Spec}A\subset X$, $f_X$ sends a $A$-module to the $A$-module given by the restriction of scalar.