For (1), recall that if $R$ is a ring, then a derivation $D: R \to R$ satisfies the Leibniz rule, which by induction on $n$ implies that if $D^n$ denotes the $n$-fold iterate of $D$, then $$D^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i(f) D^{n-i}(g).$$ Since $\binom{p}{i} \equiv 0$ for $0<i<p$, this implies that if $R$ is an $\mathbf{F}_p$-algebra, then $D^p(fg) = D^p(f) g + f D^p(g)$. In other words, $D^p$ is a derivation.
Let me answer (3) before (2). If you work locally, i.e., consider the derivation $\partial_x$ on $\mathbf{F}_p[x]$, then $(\partial_x)^p x^n$ is zero for $n<p$, and is $p! \binom{n}{p} x^{n-p}$ for $n\geq p$, which is zero. So the derivation $(\partial_x)^p$ is identically zero.
Let's now discuss (2); since everything is local on $X$ and $X$ is smooth, we can assume that $X$ is etale over $\mathbf{A}^n_S$ with basis $x_1, \cdots, x_n$. Let $\omega$ be a closed $1$-form on $X$. Because of the Cartier isomorphism $\mathfrak{C}: \mathcal{H}^i(F_\ast \Omega^\bullet_{X/S}) \xrightarrow{\sim} \Omega^i_{X^{(p)}/S}$, we can write $\omega = df + \sum_{i=1}^n F^\ast(g_i) x_i^{p-1} dx_i$ for some functions $g_i$ and $f$ on $X$. Both $\langle \mathfrak{C} \omega, D\rangle^p$ and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle$ kill $df$, so by Frobenius semilinearity, we can assume that $\omega = x_i^{p-1} dx_i$. We'll also just assume $n=1$ and write $x$ instead of $x_1$.
Then $\langle \mathfrak{C} \omega, D\rangle = \langle dx^{(p)}, D\rangle = D(x)$, and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx)$. So we need to prove that $$(Dx)^p = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx),$$ which is an identity due to Hochschild. See https://joshuamundinger.github.io/assets/notes/hochschild-identity.pdf for a cute argument; it reduces to using the multinomial analogue of the Leibniz rule for $D^p(x^p)$ and a multinomial analogue of the binomial coefficient vanishing.