[See Edit below.]
This isn't really an answer, but I believe it is relevant.
Work geometrically, so $k$ is alg. closed. Let $G$ reductive over $k$, and let
$V$ be a $G$-module (linear representation of $G$ as alg. gp.).
If $\sigma$ is a non-zero class in $H^2(G,V)$, there is a non-split extension
$E_\sigma$ of $G$ by the vector group $V$ -- a choice of 2-cocyle representing
$\sigma$ may be used to define a structure of alg. group on the variety
$G \times V$. Here "non-split" means "$E_\sigma$ has no Levi factor".
And if $H^2(G,V) = 0$, then any $E$ with reductive quotient $G$ and
unipotent radical that is $G$-isomorphic to $V$ has a Levi factor.
You can look at the $H=\operatorname{SL}_2(W_2(k))$ example from this viewpoint;
$H$ is an extension of $\operatorname{SL}_2$ by the first Frobenius twist
$A = (\mathfrak{sl}_2)^{[1]}$ of its adjoint representation. Of course, this point of view doesn't really help to see that $H$ has no Levi factor; the fact that $H^2(\operatorname{SL}_2,A)$ is non-zero only tells that it might be interesting (or rather: that there is an interesting extension).
The extension $H$ determines a class in that cohomology group, and the argument
in the pseudo-reductive book of Conrad Gabber and Prasad -- or a somewhat clunkier representation theoretic argument I gave some time back -- shows this class to be non-zero, i.e. that $H$ has no Levi factor.
So stuff you know about low degree cohomology of linear representations comes up. And this point of view can be used to give examples that don't seem to be related to Witt vectors.
A complicating issue in general is that there are actions of reductive $G$ on a product of copies of $\mathbf{G}_a$ that are not linearizable, so one's knowledge of the cohomology of linear representations of $G$ doesn't help...
Edit: It isn't clear I was correct last April about that "complicating issue". See this question.
Also: the manuscript arXiv:1007.2777 includes a "cohomological" construction
of an extension $E$ of SL$_3$ by a vector group of dim $(3/2)(p-1)(p-2)$ having no Levi factor in char. $p$, and an example of a group having Levi factors which aren't geometrically conjugate.
Much of the literature in characteristic 0 is older and may not immediately fit the exact format of your question. But here is a sample:
N. Jacobson's 1962 book Lie Algebras (later reprinted by Dover) discusses in II.5 (and II.11) the Lie algebra structure of a completely reducible linear Lie algebra in characteristic 0. His Theorem 8 shows that the Lie algebra decomposes into its center plus a semisimple ideal. (In the parallel algebraic group setting, the Lie algebra of a completely reducible linear algebraic group in characteristic 0 has this form.) The semisimple or just simple case is really the crucial one for Jacobson-Morozov theory.
In the 1968-69 IAS seminar volume published as Springer LN 131 in 1970, look at section III.4 in Conjugacy classes by T.A. Springer and R. Steinberg; here there are also adaptations to prime characteristic.
R.W. Carter's 1985 book Finite Groups of Lie Type (Chapter 5) has a nice treatment, though he usually works with simple algebraic groups.
There is also a Lie algebra treatment in section 11 of Bourbaki's Chap. 8 in the Lie groups series, supplemented by interesting exercises.
In characteristic 0 the exponential map works well to pass from the Lie algebra to the group, but in characteristic $p$ the Jacobson-Morozov argument only works for large enough $p$. For refinements involving the groups when $p$ is small, there are substantial papers by G. McNinch and D. Testerman in the
past decade or so. At any rate, the case $g=1$ or $n=0$ of your question
is trivial and can be left aside.
As BCnrd observes, you have to be careful to specify your maps to be algebraic
group homomorphisms. In characteristic 0, working with an algebraically closed field isn't so important, but in general you have to treat points of the group over a field with care. (There is a careful study by Borel and Tits of the way abstract group homomorphisms relate to algebraic group morphisms, but you don't want to get into that here.)
Best Answer
To each unipotent element $u\in G$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of $G$ are even.