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[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.

[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.

[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.

[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.

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...

[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.