Let $X$ be a group scheme over $S$, $g\in X(S)$. Then we can define the translation by point $g$. $l_g:X\to X$, $l_g=m\circ (id,g\circ *)$, where $*:X\to S $. And the morphism acts on $X(T)$ as translation. The following lemma is from stackproject https://stacks.math.columbia.edu/tag/047J
Lemma 39.7.7.
Let $k$ be a field. Let $ψ:G′→G$ be a morphism of group schemes over $k$. If $ψ(G′)$ is open in $G$, then $ψ(G′)$ is closed in $G$.
Proof. Let $U=ψ(G′)⊂G$. Let $Z=G \backslash ψ(G′)=G\backslash U$ with the reduced induced closed subscheme structure. By Lemma 39.7.2 the image of
$Z×_kG′⟶Z×_kU⟶G$ is open (the first arrow is surjective). On the other hand, since $ψ$ is a homomorphism of group schemes, the image of $Z×_kG′→G$ is contained in $Z$ (because translation by $ψ(g′)$ preserves $U$ for all points $g′$ of $G′$; small detail omitted). Hence $Z⊂G $is an open subset (although not necessarily an open subscheme). Thus $U=ψ(G′)$ is closed.
Here it seems that it has defined tranlation by any point in $U$, but I have no idea how to do that. Basically I want to know why the image of $(G\backslash U)\times_k U\to G$ is contained in $G\backslash U$, where $U$ is open subgroup of $G$.
If $\mathscr{C}$ is a category and $G$ is a group object in $\mathscr{C}$ with multiplication $m\colon G\times G\to G$, then for any morphism $f\colon A\to G$ in $\mathscr{C}$ (a generalized point of $G$) we get a generalized translation morphism $m\circ(\mathrm{id}_G\times f)\colon G\times A\to G\times G\to G$. In the special case where $A\cong *$ is a terminal object, you get a more ''classical'' translation map $G\to G$.
In case $\mathscr{C}=\mathsf{Sch}_{/k}$, then their map $Z\times_k G'\to G$ is a restriction of the generalized translation map $G\times G'\to G$ given to you by $\psi\colon G'\to G$. This is what they mean with translation by elements in $G'$. For instance, given an $R$-valued point $g'\colon\mathrm{Spec}\,R\to G'$, we get a translation map $G\times\mathrm{Spec}\,R\to G\times G'\to G$ which encodes translation by $g'$.