Let $G$ be an algebraic group over an algebraically closed field of characteristic zero $K$ and let $L$ be another algebraically closed field, together with an embedding $K \hookrightarrow L$.
Why is it true that the extension of scalars is an equivalence of categories from finite dimensional $K$-representations of $G$ to finite-dimensional $L$ representations of $G \times_K L$?
Are all the assumptions on the fields (algebraically closed, characteristic zero) needed?
First of all, even setting aside the issue with scalar automorphisms noted in comments, at the level of objects there is a "problem": the functor is not essentially surjective for unipotent groups (by a consideration of Ext$^1$-groups with their natural structure of vector space over the ground field). Probably nobody cares, so I won't get into the details (but if someone does care then hopefully someone else will summon the energy to write out an actual argument, since I don't feel like going down that road here).
Now let's focus on positive statements that someone might care about. Let $K'/K$ be an extension of fields (any characteristic, but characteristic 0 is especially nice when $G^0$ is reductive because of the Remark below), and $G$ a smooth affine group over $K$. Assume $K$ is algebraically closed (but $K'$ is not assumed to be algebraically closed). Then we claim:
Theorem: For any semisimple linear representation $\rho':G_{K'} \to {\rm{GL}}(V')$, there exists a semisimple linear representation $\rho:G \to {\rm{GL}}(V)$ unique up to isomorphism such that $\rho' \simeq \rho_{K'}$.
Remark: In characteristic 0 the semisimplicity hypothesis on the representations automatically holds if $G$ has reductive identity component. There is a version of the Theorem in characteristic 0 that doesn't require $K$ to be algebraically closed if $G$ is split connected reductive, but that involves an entirely different ingredient than is used below, so I'll pass over it in silence.
To prove the Theorem, we begin with:
Lemma The coefficients of the characteristic polynomial of $\rho'$ come from $K[G] \subset K'[G_{K'}]$.
Proof. An element $f' \in K'[G]$ comes from $K[G]$ if and only if its restriction to a Zariski-dense subset of $G(K)$ belongs to $K$. (Indeed, if we apply "spread out and specialize" to $f’$ then we get some $f \in K[G]$ such that $f_{K'}$ agrees with $f'$ on a Zariski-dense subset of $G(K)$, but such a subset is also Zariski-dense in $G_{K'}$ (exercise!), so $f' = f_{K'}$ as desired.)
Let $B$ be a Borel $K$-subgroup of $G$, so the $G(K)$-conjugates of $B(K)$ cover $G(K)$ (since $K$ is algebraically closed). Hence, it suffices to show that the coefficients on all such $G(K)$-conjugates have values in $K$. Let $T$ be a maximal $K$-torus in $B$, so $B = T \ltimes U$ for $U := \mathscr{R}_u(B)$. Applying Lie-Kolchin to $B_{K'}$ acting on $V'$, those coefficients on any $b \in B(K)$ only depend on the $T$-component of $b$. Thus, since the characteristic polynomial is conjugation-invariant, we're reduced to studying these coefficients on points of $T(K)$. All weights of $T_{K'}$ are "$K$-rational" (since $T$ is a split $K$-torus, as $K$ is algebraically closed), so we win.
QED Lemma
Now if we apply "spreading out and specialization" followed by semisimplification, from $\rho'$ we get a semisimple $$\rho: G \to {\rm{GL}}(V)$$ such that $\rho_{K'}$ has the same characteristic polynomial as $\rho'$ due to the Lemma. But $\rho_{K'}$ is semisimple because $\rho$ is semisimple with $K$ algebraically closed (i.e., $V$ is a semisimple representation of the abstract group $G(K)$ with $K$ algebraically closed, so it is "absolutely semisimple" and hence — by consideration of the endomorphism algebra — $V_{K'}$ is semisimple over $K'$ as a representation of the abstract group $G(K)$ and thus as a representation of the algebraic group $G_{K'}$ by Zariski-density of $G(K)$ in $G_{K'}$). Hence, $\rho_{K'}$ and $\rho'$ are semisimple representations of $G(K')$ with the same characteristic polynomial, so by Brauer-Nesbitt (which applies to semisimple representations of finite dimension for any abstract group at all) these representations are isomorphic. That isomorphism amounts to a single conjugation in the language of matrices, so it says that as algebraic representations they’re $K'$-isomorphic. This gives that $\rho'$ descends to a semisimple representations of $G$ over $K$ as desired.
QED Theorem
This actually strikes me as a question about the foundations chosen for algebraic geometry, which in turn affect formlulations about linear/affine algebraic groups and their representations. It's useful to recall that the foundations were in flux for a long time, while people like Weil experimented with working over a very big "universal domain" with an unlimited supply of indeterminates. Eventually Grothendieck's school developed a more flexible language for talking about "schemes" (including algebraic groups) over arbitrary commutative base rings including fields which may or may not be algebraically closed.
In the 1960s Borel adapted the scheme theory in an ad hoc way to reach the structure theory of semisimple (or reductive) groups more quickly: Bruhat decomposition, parabolic subgroups, etc. Anyway, there is general agreement now that the choice of an algebraically closed field won't affect significantly the properties of a scheme/variety or in particular of an algebraic group. This applies equally well to the (finite dimensional) representations which are "rational" (algebraic). So an extension involving just algebraically closed fields leaves all the basic data intact. See for example Milne's recently published book on algebraic groups here, though it's still quite a long story when told in the "correct" language of algebraic geometry.
