This is not a direct answer to the question for a general group scheme $G \to S$ and I am not an expert in this area. However, I would like to point out that the resolution property of stacks is a natural condition that appears in this context of Hilbert's 14th problem by work of R. W. Thomason:
Equivariant resolution, linearization, and Hilbert's fourteenth problem over arbitrary base schemes
Advances in Mathematics 65, 16-34 (1987)
Once and for all let $ \pi \colon G \to S$ be an affine, flat, finite type group scheme over a noetherian and separated base scheme $S$.
Recall, that a noetherian algebraic stack has the resolution property if every coherent sheaf is a quotient of a vector bundle (a locally free sheaf, which will be always assumed to be of finite and constant rank).
Therefore, the classifying stack $B_S G$ has the resolution property if and only if every coherent $G$-comodule on $S$ is the equivariant quotient of some locally free $G$-comodule. The latter is the definition of the $G$-equivariant resolution property of $S$.
What we need is his Theorem 3.1:
$G \to S$ can be embedded as a closed subgroup scheme of $GL(V)$ for some vector bundle $V$ on $S$ if $B_S G$ has the resolution property. If $S$ is affine, $V$ can be taken to be free.
Thomason does not say that the converse to Theorem 3.1. also holds. I guess that this is true if $S$ is affine, but as I am always getting confused while working with comodules, I cannot give a rigorous proof at the moment.
Nevertheless, it is worth to ask when $B_S G$ has the resolution property. Thomason proved this in the following cases:
- $S$ regular and dim $S \leq 1$,
- $S$ regular; dim $S \leq 2$; $\pi_* O_G$ is a locally projective $O_S$-module, e.g, if $\pi \colon G \to S$ is smooth and with connected fibres.
- $S$ regular or affine or has an ample family of line bundles; $G$ a reductive group scheme which is either split reductive, or semisimple, or with isotrivial radical and coradical, or over a normal base $S$.
In particular, if $S$ is the the spectrum of the ring of dual numbers, then this provides an affirmative answer to the posted question if $G \to S$ satisfies the conditions in (3).
Even for $G \to S$ arbitrary with reduction $G_0 \to S_0$, we know that the reduction $X_0=B_{S_0}G_0$ of $X= B_S G$ has the resolution property by (1).
So we may reformulate the original question as follows:
(Q2) Is the resolution property preserved under the first order deformation $X_0 \to X$?
Lifting of various locally free resolutions from $X_0$ to $X$ is probably not the best approach. However, it suffices to lift a single locally free sheaf.
Let us see, why this is true.
A noetherian algebraic stack with affine diagonal has the resolution property if and only if there exists a vector bundle $V$ whose associated frame bundle has quasi-affine total space.
The normal case was proven by Totaro in
The resolution property for schemes and stacks.
J. Reine Angew. Math. 577 (2004), 1--22.
14A20 (14C35)
and in my thesis, I am currently working on, I show that this really holds for non-reduced stacks too.
Therefore if we can lift $V_0$ from $X_0$ to a vector bundle $V$ on $X$, then $V$ has still quasi-affine frame bundle as its reduction is quasi-affine.
The obstruction for this lies in $H^2(X_0, I \otimes V_0^\vee \otimes V_0)$ where $I$ is the coherent ideal of order two defining the deformation $X_0 \to X$. Probably, the ideal can be removed here with some tricks.
In our case this cohomology boils down to the second group cohomology of the $G_0$-representation $I \otimes V_0^\vee \otimes V_0$. In particular, if $G_0 \to S_0$ is linearly reductive, the obstruction is zero.
Therefore we have proven:
If $G \to S$ is a group scheme over an artinian base with linearly reductive special fibre, then $G \to S$ can be embedded into some $GL_{n,S}$ as a closed subgroup scheme.
Clearly this still leaves out interesting cases and probably this can be proven more directly avoiding stack theory.
Here are some considerations on the case $X$ smooth.
Let $d$ be the degree of $X$ and let $L$ be the restriction to $X$ of $O_{P^2}(1)$.
If $k=1$ then the condition is precisely that the line bundle $L(-dP)$ is a torsion point of $Pic^0(X)$. In fact let $m$ be such that $mL(-dP)$ is trivial. Since the map $H^0(P^2,{\cal O}_{P^2}(m))\to H^0(X, mL)$ is onto, there exists a curve $Y$ of degree $m$ that intersects $X$ precisely at $P$ with multiplicity $md$.
So the condition is satisfied for at most countably many points $P\in X$, unless $X$ is rational.
One can argue in a similar (more complicated) way for $k>1$.
I don't know if the remark that follows is useful.
If $X$ is smooth of genus $g$, $P\in X$ is fixed and $k=g+1$, then one can consider the image of the map $X^g\to Pic^0(X)$ defined by mapping $(P_1,...,P_g)$ to $(g+1)L(-d(P+P_1+...+P_g))$. This map is surjective, so the above argument implies that, given $P$, one can find $g$ points such that there exists a curve $Y$ that intersects $X$ only at $P, P_1,\dots P_{g}$. Since the subvarieties {P_i=P} of $X^g$ and the weak diagonal map to proper subvarieties of $Pic^0(X)$ and the torsion points are dense in $Pic^0(X)$, one can find $P,P_1,...P_g$ distinct.
More generally, if $k>g$ one can assign $k-g$ points and find a curve $Y$ that meets $X$ at those points and at precisely $g$ additional points.
Best Answer
You make your solution feel "fancier", you could start with a more coordinate free approach to the dual curve. For example, try to identify the line bundle on your curve which embeds it into the dual plane. E.g.: let $X \subset \mathbb{P}^2$ be your curve. Then define $X' \subset X \times \mathbb{P}^{2*}$ to be the set of pairs $(x, H)$ where the line $H$ is contained in the tangent space to $X$ at $x$. When $X$ is smooth at least, then this may be identified with the projectivization of the dual of the normal bundle $N_{X/\mathbb{P}^2}$.
Then think about the map $X' \rightarrow X \times \mathbb{P}^{2*} \rightarrow \mathbb{P}^{2*}$. This is the dual curve embedding. The morphism comes from the inclusion of $N^* \subset \Omega_{\mathbb{P}^2}|_X \subset \mathcal{O}(-1)^3$.
At some point, you have to do computations in local coordinates - but setting it up like this may help a little.
Example of local computation: Locally, the curve looks like $(x(t), y(t), z(t))$. The map to the dual plane is given by $(x(t), y(t), z(t)) \times (x'(t), y'(t), z'(t))$ (cross product! giving the normal to the tangent plane in $\mathbb{C}^3$). If the curve has a simple inflection point, then locally it looks like $(t, t^3, 1)$ and the map to the dual plane is given by $(3t^2, 1, 2t^3)$ - a curve with a simple cusp.