Let $F \subseteq k$ be a finite purely inseparable field extension with degree $p^d$, $p = \textrm{Char } k$. If $x_1, ... , x_d \in k$ with $k = F[x_1, ... , x_d]$, it follows that we have an isomorphism of $F$-algebras
$$F[T_1, ... , T_d]/(T_1^{n_1}-x_1, ... , T_d^{n_d}-x_d) \rightarrow k$$
Partial differentiation with respect to $i$ is well defined on the quotient of $F[T_1, ... T_d]$, which induces an $F$-derivation $\partial_i$ of $k$. Now if I'm understanding correctly, $\mathscr J = \textrm{Der}_F (k,k)$ is a $p$-Lie algebra with $p$-operation $\delta^{[p]} = \delta \circ \cdots \circ \delta$ ($p$ times). The book I'm reading (Springer, Linear Algebraic Groups, 11.1.13) claims that $\partial_i^{[p]} = \partial_i$ for all $i$.
This doesn't make sense to me, since we should have $\partial_i(x_i) = 1$, but $\partial_i^{[p]}(x_i) = 0$.
Careful, you have a few things slightly off. $\phi : F[T_1,\dots, T_d]/(T_1^p - a_1,\dots, T_d^p - a_d)\to k$ is the isomorphism, where the $a_i$ live in $F$, not $k$. (Well, they do live in $k$, but they are also in $F$, which is the important bit: $k = F[x_1,\dots, x_d] = F[a_1^{1/p},\dots, a_d^{1/p}]$. What you have written doesn't make sense if the $x_i$ are generators of $k/F$, since $T_i^p - x_i\not\in F[T_i]$.)
More importantly, what appears to be the source of your confusion is that the derivation Springer is looking at is not induced by $D_i$ (partial differentiation with respect to $i$), but rather by $T_iD_i$: so if $\partial_i$ is the induced derivation, then $\partial_i(x_j) = \phi(T_iD_i(T_j)) = \phi(\delta_{ij}T_i) = \delta_{ij}x_i$, where $\delta_{ij}$ is $1$ if $i = j$ and $0$ otherwise. It is easy to see that this $\partial_i$ indeed satisfies $\partial_i^{[p]} = \partial_i$.