Let $G$ be a unipotent connected linear algebraic group over a field $F$. Then $G$ is called split if there is a series of closed subgroup schemes $1 = G_0 \subset G_1 \subset \cdots \subset G_t =G$ with each $G_i$ normal in $G_{i+1}$ and $G_i/G_{i+1} \cong_F \mathbb G_a$.
I have read that for $F$ perfect, every such $G$ is split, provided it is connected. And $F$ of characteristic zero, every such $G$ is moreover automatically connected.
I was trying to verify this in the case of $G = \operatorname{Res}_{E/F} \mathbb G_a$, where $E/F$ is a quadratic extension. Does $G$ really have such a composition series?
I tried considering the diagonal embedding $\Delta$ of $\mathbb G_a$ into $G$ and looking at the quotient $G/\Delta$. It should be the case that $G/\Delta \cong \mathbb G_a$, but I'm not able to verify this. The natural map $G/\Delta \rightarrow \mathbb G_a$ given by $(x,y)\Delta \mapsto x- y $ is not defined over $F$.
It's not that complicated. I believe you have all the needed knowledge, but just have to unpack the definitions.
By fixing a basis of $E/F$, we know that $E$ is isomorphic to $F^2$ as $F$-vector spaces, hence for any $F$-algebra $R$, one has $\mathbb{G}_a(E\otimes_F R) = E\otimes_F R \simeq R^2 = \mathbb{G}_a^2(F)$, where the middle isomorphism is functorial in $R$.
Hence by the definition of restriction of scalars, we see that $\operatorname{Res}_{E/F}\mathbb{G}_a \simeq \mathbb{G}_a^2$.