Since you ask about other situations where this sort of thing occurs, let me describe a general principle (applied to the context of the original question) which is widely applied in EGA and elsewhere, often called "spreading out and clever specialization". It is not as efficient as Torsten's argument, and may look quite long at first glance, but hopefully by the end you'll see that it conveys a quite simple idea that is broadly useful for this kind of issue.  In particular, it doesn't use the finer information about local closedness of orbits (simple as that may be), so it is applicable to contexts way beyond the one of group 
actions. 

I will work with quite weak hypotheses to emphasize the general applicability and flexibility of the basic idea. Then you'll also see that in a discussion between two experts, this would all be disposed of in a few sentences (so the length of what follows may create the wrong impression about the complexity).

Setup: Suppose a finite type group scheme $G$ over an algebraically closed field $k$ acts on a finite type $k$-scheme $X$ (assume $G$ and $X$ are affine if you wish), and $K$ is a nonzero $k$-algebra (perhaps not an algebraically closed field). I claim $j:X(k)/G(k) \rightarrow X(K)/G(K)$ is injective (so the source is finite whenever the target is finite), and (more interestingly) that it is also surjective if $X(k)/G(k)$ is finite and $K$ is an algebraically closed field. That will answer the original equivalence question.

First let's do injectivity.  Since $K$ exhausted by finite type $k$-subalgebras $K_i$ (definitely *not* fields in general), we have $X(K)= \varinjlim X(K_i)$ and $G(K)= \varinjlim G(K_i)$ (as $X$ and $G$  are finite type, or alternatively it is clear in the affine case).  Thus, $X(K)/G(K) = \varinjlim X(K_i)/G(K_i)$, so it enough to treat the $K_i$ in place of $K$. So we can assume $K$ is finitely generated as a $k$-algebra. [This is a powerful idea, even when the original $K$ is a field.] By the Nullstellensatz there is a $k$-algebra map $s:K \rightarrow k$ (quotient by any maximal ideal) with $k \rightarrow K$ as section; this is the "specialization" trick. It defines a map of sets $X(K)/G(K) \rightarrow X(k)/G(k)$ with the original map $j$ as a section (as $A \rightsquigarrow X(A)/G(A)$ is a functor on $k$-algebras $A$), so $j$ is injective. 

 That was a more or less a formal kind of silliness (despite neat use of the Nullstellensatz), so now we come to the interesting part: assuming $K$ is an algebraically closed field and $X(K)/G(K)$ has at least $n$ points then so does $X(k)/G(k)$ (so surjectivity follows when $X(k)/G(k)$ is finite). The basic principle is this:

> whatever finite amount of stuff happens over an algebraically closed extension of an algebraically closed field already happens over the ground field via well-chosen specialization.  (Kind of like those ads about Las Vegas.)

Say $x_1,\dots,x_n$ in $X(K)$ lie in distinct orbits. Exhausting $K$ by finitely generated $k$-subalgebras $K_i$ as above, we can find a big enough $K_i$, call it $A$, so that $x_1,\dots,x_n \in X(A)$.  We want to show that for a "sufficiently generic" specialization map $A \rightarrow k$, their images in $X(k)/G(k)$ remains distinct. Here, the valuable geometric intuition (which makes sense even within classical algebraic geometry, since $k$ is algebraically closed and $S :=$ Spec($A$) is basically a classical irreducible variety, as $A$ is a domain of finite type over $k$) is that the $x_i \in X(A)$ are sections to the projection $X \times S \rightarrow S$ such that on the *geometric* generic fiber over $S$ (i.e., pullbacks along $A \rightarrow K$) they are is pairwise distinct $G$-orbits, and we want to claim that under specialization over some dense open in $S$ they remain in pairwise distinct $G$-orbits.

Since any intersection of finitely many non-empty opens in the irreducible $S$ contains a $k$-point (Nullstellensatz once again), it suffices to prove a more general fact for a pair of points $x, x' \in X(A)$ (to then be applied to each of the *finitely many* pairs $x_i, x_{i'} \in X(A)$ with $i \ne i'$): I claim that if their images in $X(K)$ (the "geometric generic fiber") are in distinct $G(K)$-orbits, then there's a dense open $U$ in $S$ such that for any $u \in U$ (e.g., a $k$-point!) the specializations $x(u), x'(u) \in X(k(u))$ have disjoint orbits under the action of $G_{k(u)}$ on $X_{k(u)}$ (in the sense of their orbit subvarieties over $k(u)$, or geometric points thereof, which comes to the same thing).  This will clearly do the job.

OK, now comes the step where we make an actual group-theoretic construction (akin to what Torsten did more efficiently) to produce the required open: we view $G \times S$ as an $S$-group acting on the $S$-scheme $X \times S$ and form a "transporter scheme".  That is, consider the closed subscheme 
$$T_{x,x'} = \{g \in G \times S\,|\,g(x) = x'\} \subset G \times S$$ 
over $S$. In more precise terms, writing $G_S$ and $X_S$ as shorthand for $G \times S$ and $X \times S$ to save space, we have the action map $G_S \rightarrow X_S$ over $S$ defined functorially by $g \mapsto g.x$, and $T_{x,x'}$ is the preimage of the closed subscheme  of the target given by the (closed immersion) section $x':S \rightarrow X_S$ over $S$. 
For example, if $s \in S(F)$ for a field $F/k$ then the $s$-fiber of $T_{x,x'}$ is the closed subscheme of $G_F$ defined by the condition "$g.x(s) = x'(s)$" for the points $x(s), x'(s) \in X(F)$. In effect, $T_{x,x'}$ is just the relative version of this latter classical transporter construction as we vary across the pairs $(x(s),x'(s))$ for $s$ wandering in $S$.

Finally we have assembled enough to finish. Consider the structural morphism $q:T_{x,x'} \rightarrow S$.  This is a map between finite type $k$-schemes. What is its fiber (i.e., pullback) over a point $s:{\rm{Spec}}(F) \rightarrow S$ (such as a $k$-point, or more importantly the "geometric generic point" ${\rm{Spec}}(K) \rightarrow S$)? Well, we just saw what this is: it is the "classical" transporter for $x(s), x'(s) \in X(F)$ inside of $G_F$. So the fiber of $q$ over a physical point $s \in S$ is *empty* precisely when the corresponding transporter (a finite type $k(s)$-scheme) is empty, which is to say that $x(s), x'(s) \in X(k(s))$ have disjoint orbits under $G_{k(s)}$ acting on $X_{k(s)}$ (i.e., in distinct orbits under $G(\overline{k(s)})$ acting on $X(\overline{k(s)})$, not merely under $G(k(s))$ acting on $X(k(s))$, since emptiness of a finite type $k(s)$-scheme amounts to the absence of $\overline{k(s)}$-points and not merely of $k(s)$-points).  Excellent, so if the image of $q:T_{x,x'} \rightarrow S$ misses a dense open $U$, that open will do the job (i.e., for all $u \in U$, the points $x(u), x'(u) \in X(k(u))$ lie in distinct $G_{k(u)}$-orbits in $X_{k(u)}$).  Aha, but by (the scheme version of!) Chevalley we know that the image of $q$ is a *constructible* set even at the level of schemes, so if it misses the generic point then it misses a dense open as desired.  So we are reduced to proving that $q$ has empty fiber over the generic point of $S$.  But that in turn is exactly the original hypothesis that on the *geometric* generic fiber over ${\rm{Spec}}(K) \rightarrow S$ our points $x, x' \in X(K)$ lie in distinct orbits under the $G(K)$-action. Voila. QED

Now you can see the one serious ingredient that uses schemes (going beyond classical algebraic geometry) in an essential way: the validity of Chevalley's theorem on constructible images in the scheme framework, and the ability to apply it in conjunction with the literal generic point (and geometric points over that).  Hopefully you can see that (together with specialization) this is a broadly useful technique for propogating results from an algebraically closed extension of an algebraically closed field back down to the ground field.  And that once one realizes this idea, it is sort of simple in the end.