Recall that if $G\rightarrow S$ is a flat group scheme, then a $G$-torsor is an
$S$-scheme $X\rightarrow S$ with a $G$-action $G\times_SX\rightarrow X$ such that
$G\times_SX\rightarrow X\times_SX$ given by $(g,x)\mapsto (gx,x)$ is an
isomorphism <em>and</em> such that $X\rightarrow S$ is faithfully flat. This
already gets rid of the flat topology but in the current case where
$S=\mathrm{Spec} R$ a local Artinian ring and we also give ourselves an
isomorphism of $G$-schemes $X\times_S\mathrm{k}=G\times_S\mathrm{Spec}k$ then it
is enough that $X\rightarrow S$ is a flat non-empty $G$-scheme as the fact that
$G\times_SX\rightarrow X\times_SX$ is an isomorphism can be checked upon
reduction to $k$.

As then $X$ is affine we are talking about a finite flat (non-zero) $R$-algebra
$T=R[X]$ which is a $R[G]$-comodule which is also a comodule algebra (i.e., the
product $T\bigotimes_RT\rightarrow T$ as well as the unit $R\rightarrow T$
are comodule maps) together with a comodule algebra isomorphism
$T\bigotimes_Rk\rightarrow k[G]$. If we specialise further to the actual case at
hand, the category of $U({\frak g}\bigotimes_kR)^\ast$-comodules is isomorphic (for
once this is really an isomorphism) as a tensor category to the category of
${\frak g}\bigotimes_kR$-modules (as $p$-Lie algebra). This gives us a description purely in terms of
${\frak g}\bigotimes_kR$-modules.

Note that there is also a very concrete description of $G$-torsors (for
$G\rightarrow S$ finite flat for simplicity over an affine $S=\mathrm{Spec}R$).
For an $R$-algebra $R'$ we have a tautological map of group schemes
$G(R')\hookrightarrow (R'[G]^\ast)^\times$ (an element $f\in G(R')$ is by
definition an $R'$-homomorphism $R'[G]\to R'$ i.e. an element of $R'[G]^\ast$, it is easily seen to land in
$(R'[G]^\ast)^\times$ and is tautologically a group homomorphism). This gives us
an embedding of flat group schemes $G\hookrightarrow (R[G]^\ast)^\times$. This
gives us a (half-)long exact sequence of cohomology sets associated to
$G\hookrightarrow (R[G]^\ast)^\times\rightarrow (R[G]^\ast)^\times/G$. If $R$ is
Artinian all $(R[G]^\ast)^\times$-torsors are trivial as they correspond to
locally free rank $1$-modules over $R[G]^\ast$ (right modules to be specific)
which are all trivial as $R[G]^\ast$ is also Artinian. Hence  $G$-torsors
correspond to sections of $(R[G]^\ast)^\times/G$ modulo the action of the
section of $(R[G]^\ast)^\times$. However, it is in general difficult to get a
concrete description of $(R[G]^\ast)^\times$ and even when you have one the
orbits may be difficult to figure out.

<b>Addendum</b>: In the orbit description I forgot to add the condition that everything should map to the identity in $k$. Note also that the orbit description is no doubt the closest you can get in general to usual description of for instance $\alpha_p$- and $\mu_p$-torsors where one extracts a $p$-th root of a function (resp. an invertible function). In some other cases one also gets simpler descriptions. For instance if $H$ is a smooth $R$-group scheme and $G$ is the kernel of the Frobenius map, then every $G$-torsor with a trivialisation over $k$ is obtained from a section of $H^{(p)}$.

<b>Addendum 1</b>: As an answer to Jacob's followup question, tautologically the image of $G$ in $(R[G]^\ast)^\times$ consists of the elements $f$ for which $\Delta(f)=f\otimes f$, where $\Delta$ is the coproduct. This means (I hope) that if we let $V$ be the graph of $\Delta$ as a subspace of $R[G]^\ast\times R[G]^\ast\bigotimes [G]^\ast$, then $G$ is the stabiliser of $V$. From this we can get a $1$-dimensional subspace by taking the $\dim V$-th exterior power. Mind you this gives a $1$-dimensional subspace not a vector. In many cases one does however get the same result when looking at a fixed vector.