Let $R$ be a complete noetherian local ring and $G$ be a finite group scheme over $R$ (that is a group scheme over $\operatorname{Spec}(R)$ whose structure morphism is finite locally free, or equivalently here, finite flat). As detailed in the notes "finite flat group schemes" by Tate, there is a canonical exact sequence $$0\rightarrow G^0 \rightarrow G\rightarrow G^{ét}\rightarrow 0$$ where $G^0$ is the identity component of $G$, it is a finite connected group over $R$, and $G^{ét}$ is the quotient, it is a finite étale group over $R$.
In the other paper "p-divisible group", Tate states that in general, we have an identification $G^{ét}=G(\overline{k})$ where $\overline{k}$ is an algebraic closure of the residue field $k$ of $R$.
I am confused by this last statement. First of all, $G^{ét}$ is a group scheme over $R$ whereas $G(\overline{k})$ is, to my knowledge, just an abstract group whence I don't know how to interpret this equality.
Moreover, in all generality if $X$ is a scheme defined over a basis ring $A$ and $L$ is an $A$-algebra which is also a field, then the $L$-valued points $X(L)$ of $X$ are in natural identification with pairs $(x,\phi)$ where $x\in |X|$ and $\phi: \kappa(x)\rightarrow L$ is a field extension, $\kappa(x)$ being the residue field of $x$.
Now if $X$ is a group scheme $G$, it is a homogeneous space over itself by translation, hence all points of $G$ have the same local ring, so the same residue field as well. Whence, if $G$ has an $L$-valued point, then all its points are $L$-valued.
If I take this observation into account (and if it is true), then the identity $G^{ét}=G(\overline k)$ would imply $G^{ét}=G$ or $G^{ét}$ is trivial according to whether there is a $\overline{k}$-valued point or not. This conclusion does not sound right to me, hence there must be a mistake in my understanding.
Could somebody please help me clarify the situation here ?