The key point is to show the following: let $f:V\rightarrow W$ be a regular map of varieties over an algebraically closed field $k$; if $V(k)\rightarrow W(k)$ is surjective, then $V(K)\rightarrow W(K)$ is surjective for every algebraically closed field $K$ containing $k$.
We prove this by induction on the dimension of $W$. We may suppose that $W$ is an irreducible closed subvariety of some affine space. Let $P\in W(K)$ be not in the image. We know that $tr.deg.k(P)\leq dimW$. If equality holds, then $P$ and its conjugates under $Aut(K/k)$ are Zariski dense in $W_{K}$, contradicting the fact that $f$ is (obviously) dominant. Hence $P\in Z(K)$ for some proper closed irreducible subvariety $Z$ of $W$. Now apply induction to $f^{-1}(Z)\rightarrow Z$ to get a contradiction.
Edited: I add the rest of the argument. No hypotheses on $G$ are needed.
We prove: Let $G\times V\rightarrow V$ be an action of the group variety $G$ on the variety $V$, and let $K$ be an algebraically closed field containing $k$. Then $G$ has finitely many orbits on $V$ if and only if $G_{K}$ has finitely many orbits on $V_{K}$, in which case the numbers of orbits are the same and each $K$-orbit has a $k$-point.
For the proof, we first remark that if $v_{1},v_{2}\in V(k)$ lie in distinct $G$-orbits, then they lie in distinct $G_{K}$-orbits. To see this, let $Z$ be the inverse image of $v_{2}$ under the regular map $g\mapsto gv_{1}\colon G\rightarrow V$. Then $Z(k)$ is empty if and only if $Z$ is the empty variety if and only if $Z(K)$ is empty.
Suppose that $G$ has only finitely many orbits on $V$, and let $v_{1} ,\ldots,v_{m}\in V(k)$ represent the different orbits. The regular map $(g,v_{i})\mapsto gv_{i}\colon G(k)\times\{v_{1},\ldots,v_{m}\}\rightarrow V(k)$ is surjective, and hence remains surjective with $K$ for $k$. Together with the first remark, this shows that $v_{1},\ldots,v_{m}$ represent the different orbits of $G_{K}$ on $V_{K}$.
Finally, suppose that $G_{K}$ has only finitely many orbits on $V_{K}$. Then the first remark shows that $G$ has only finitely many orbits on $V$, and the previous argument applies.