Let $k$ be a field; I'm going to discuss linear algebraic groups over $k$. The question I'll pose is only interesting when the characteristic is $p>0$.
1. Some motivation
A vector group is an algebraic group isomorphic (over $k$) to a product of (finitely many) copies of the additive group $\mathbf{G}_a$. Let $U$ be any connected unipotent group over $k$. If $k$ is perfect, or if $U$ is $k$-split, then $U$ has a filtration
$1 = U_0 \subset U_1 \subset \cdots \subset U_n = U$
where each $U_i$ is normal in $U$ and each $U_i/U_{i-1}$ is a vector group. If $U$ is a normal subgroup of a linear group $G$, you can
arrange that each $U_i$ is invariant under conjugation by $G$. This suggests that to study the group extension $$(*) \quad 1 \to U \to G \to G/U \to 1,$$ one might profitably study first the case where $U$ is a vector group.
Let $G$ be a linear group and suppose that the unipotent radical $R$
of $G$ is defined over $k$ and is $k$-split (each of these conditions
can fail in general; they always hold when $k$ is perfect). Then the
question of whether $(*)$ splits when $U=R$ is precisely the question
of whether $G$ has a Levi factor; cf. this question of Jim Humphreys.
2. Action on a vector group
Let $U$ be a vector group and suppose that the linear group $G$ acts
on $U$ by algebraic group automorphisms.
The action of $G$ on $U$ determines an action of $G$ on
$\mathfrak{u}=\operatorname{Lie}(U)$.
Question (first approximation): Is there a $G$-equivariant
isomorphism $\mathfrak{u} \to U$ (where the vector space
$\mathfrak{u}$ is viewed as a vector group in the obvious fashion)?
I'll say that the action of $G$ on $U$ is linearizable if
there is such an equivariant isomorphism.
Some remarks: If the action of $G$ on $U$ is linearizable, then $G$
centralizes the action of the multiplicative group $\mathbf{G}_m$ on
$U$ obtained by transport of structure from scalar multiplication on
$\mathfrak{u}$. This $\mathbf{G}_m$-action determines a grading on
the algebra $k[U]$ of regular functions on $U$ which is stable
for the action of $G$ on $k[U]$.
3. Partial answers
Postive: If the characteristic of
$k$ is $0$, the above question has always an affirmative answer. (Use the exponential mapping $\mathfrak{u} \to U$; this is $G$ equivariant e.g. because there is a faithful linear representation of the semidirect product $G \ltimes U$).
Negative: If $G = \mathbf{G}_a$, the
question has a negative answer in char. $p>0$. Consider the action of
$G$ on $V = {\mathbf{G}_a}^2$ given by the rule $$x.(a,b) = (a +
xb^p,b).$$
The action of $G$ on $\operatorname{Lie}(V)$ is trivial, so there is no
$G$-equivariant isomorphism $\operatorname{Lie}(V) \to V$.
4. What I meant to ask.
Refined question: If $G$ is reductive, is the action of $G$ on a vector group $U$
always linearizable?
An affirmative answer would mean that when $G/U$ is reductive, one can study the group extension $(*)$ using cohomology of linear representations of $G/U$.
Note that are known examples even in char. 0 (Schwarz, Kraft,…) of
actions of reductive $G$ on
affine space $A=\mathbf{A}^n$ which are not linear, but, at least in
char. 0, these actions don't respect a vector group structure on $A$.
Best Answer
[Update below: I think this answers the question now.] Let's suppose that we're working over an algebraically closed field $k$ of characteristic $p$. Let $Fr$ denote the Frobenius automorphism.
From a recent paper of Tanaka and Kaneta (Hiroshima Math. J. 2008), I found that $Aut(G_a^n)$ (the automorphism group of the vector group $G_a^n$ over $k$) can be identified with: $$\{ A \in M_n(k)[[F]]^\times : A,A^{-1} \in M_n(k)[F] \}.$$ Here, we work in a ring of noncommutative power series in the formal variable $F$, with coefficients in the matrix ring $M_n(k)$, subject to the natural relation: $$F \cdot m = Fr(m) \cdot F.$$ I.e., one may pass all F's to the right, by applying Frobenius to the entries of the matrices.
Now, there is a natural surjective homomorphism $lin$ from $Aut(G_a^n)$ to $GL_n(k)$, obtained by taking the "constant term" of the power series.
What does the kernel of $lin$ look like? Is it (representable by?) a pro-unipotent group scheme over $k$? I have no idea.
Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.
The image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.
Update: Let $\alpha: G \rightarrow Aut(G_a^n)$ be an algebraic action of $G$ on the vector group $G_a^n$.
It follows that $H = \alpha(G)$ is a reductive subgroup of the pro-affine group $Aut(G_a^n)$, which does not intersect the pro-unipotent subgroup $U = Ker(lin)$. It remains to prove that $H$ is conjugate to the subgroup $GL_n \subset Aut(G_a^n)$.
Let $H' = lin(H)$ be the projection of $H$ in $GL_n$ -- this projection is an isomorphism from $H$ to $H'$. We find that $H \cdot U = H' \cdot U$, and $U$ is the unipotent radical of $H \cdot U$.
Now the remaining question: are Levi factors of $H \cdot U$ (geometrically, since we work over $k = \bar k$) conjugate? To this, I think I defer back to the OP, who has recently written a paper on this topic, which will hopefully adapt to this pro-affine group setting.