[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.
One method for computing branching rules in favorable situations is to use the Littelmann path model---this has a wiki page
http://en.wikipedia.org/wiki/Littelmann_path_model
In this situation (of semisimple $G$ and with $H$ the Levi subgroup of a parabolic) irreducibles essentially never remain irreducible.
Edit in response to the comment below:
Another situation that sometimes occurs (e.g. $SO_n \subseteq SL_n$) is that $H$ is the fixed point set of an involution $\phi$ of $G$. In this case, given any irreducible $G$ module $V$ we can construct another $G$-module $V^\phi$ by twisting by $\phi$. If $V$ and $V^\phi$ are isomorphic as $G$-modules, then we obtain an involution of $V$ as an $H$-module whose $\pm 1$ eigenspaces are $H$-submodules. This allows one to prove that certain $G$-modules are not irreducible as $H$-modules.
Best Answer
The short answer is that the characteristic $p$ picture genuinely has more depth. In particular it's not correct to think of formal groups as "better" than Lie groups. In fact, they can be put on equal footing with each other.
I think it's helpful to put all the objects you're interested in in the same category. In this case it is the category of group objects in formal schemes, which I will call formal group objects. There is some notational confusion here. By rights, the category of formal group schemes should denote this category (which includes all algebraic groups as well). However, often when people say "formal group scheme" or "formal group" they mean infinitesimal groups, which are formal group schemes with a single point (and other times they mean something else more restrictive, such as formally smooth infinitesimal group schemes, which I don't want to consider separately). Here I'll hopefully avoid the notational confusion and use group objects for group objects in formal schemes and infinitesimal groups for group objects with one point.
Now in characteristic zero, every algebraic group $G$ has a unique infinitesimal subgroup $\hat{G}\subset G$ (as a group object) subject to the condition of having the same tangent space, i.e., the natural map $T_e\hat{G}\to T_e G$ being an isomorphism. In characteristic $p$, this is no longer the case. If $G$ is an algebraic group in characteristic $p$ then there is still a maximal sub-group object $\hat{G}\subset G$ corresponding to the "full" formal group of $G$, but there is another smaller object $$\hat{G}^{(1)}\subset \hat{G},$$ the first Frobenius neighborhood, which also has the same tangent space (there are also objects $\hat{G}^{(2)}, \dots,$ in addition to possible other intermediate subgroups). The group $\hat{G}^{(1)}$ has the nice property that its representations are equivalent to representations of the Lie algebra $\mathfrak{g}$ as a symmetric monoidal category, and this uniquely characterizes $\hat{G}^{(1)}.$ In fact if $\mathfrak{g}$ is an arbitrary Lie algebra in characteristic $p$, it might not integrate to a smooth algebraic group $G$, but it will always integrate to a group object $G^{(1)}$ that looks like a first Frobenius neighborhood.
Now even if you look at the "full" formal neighborhood $\hat{G}\subset G,$ you will not have anything resembling integration of representations. Indeed, if $G\to \text{End}(V)$ is a representation of a smooth algebraic group scheme, you do get induced representations of $\hat{G}$ and $\hat{G}^{(1)}$ (equivalently, $\mathfrak{g}$). But you can neither integrate representations of $\hat{G}^{(1)}$ to representations of $\hat{G}$ nor representations of $\hat{G}$ to representations of $G$ in any reasonable sense, even in one-dimensional cases.