This is quite trivial but not enough for a comment: $F^G$ is not always $K^\mathbb N/\mathcal U$.

Let $x_1,x_2,\dots$ be elements of $K$ such that no product of finitely many of them is a perfect square in $K$. For instance, if $K=\mathbb Q$ we can take them to be the primes.

Our element of $\bar{K}^\mathbb N/\mathcal U$ will come from the following sequence of elements of $\bar{K}$: If $n$ is written in base $2$ as $2^{e_1}+2^{e_2}+\dots + 2^{e_k}$, the $n$th element of the sequence is $\sqrt{x_{e_1}x_{e_2} \dots x_{e_k}}$.

For each finitely generated subgroup of $G$, infinitely many elements of the sequence are invariant under it. This is because any finite codimension subspace of $\mathbb F_2^n$ stil has infinitely many elements, by induction on the codimension.

Thus, the set of all subsets of $\mathbb N$ containing the elements of the sequence invariant under any finitely generated subgroup is a filter, which can be extended to an ultrafilter.

Since for each element of $G$, the subsequence invariant under that element is in the ultrafilter, the associated element of $F$ is $G$-invariant. It lives in an algebraic extension of $G$.

Conversely, if $G$ is topologically finitely generated, then $F^G$ is $K^\mathbb N/\mathcal U$, because being if the set invariant under each generator is in the ultrafilter, the set invariant under all the generators is in the ultrafilter, and that's also the set of elements of $K$.