$\newcommand{\Gal}{\operatorname{Gal}}$If $E/F$ is a finite Galois extension, then $\Gal(\prod_UE/\prod_UF)$ is canonically isomorphic to $\Gal(E/F)$. Indeed, by the primitive element theorem, $E=F(\alpha)$ for some $\alpha\in E$. This means every element of $E$ can be written as a polynomial in $\alpha$ with coefficients in $F$ of degree smaller than $d=\deg\alpha$.

Let $\alpha^*$ be the image of $\alpha$ in the ultrapower. Then we have $\prod_UE=(\prod_UF)(\alpha^*)$, as is straightforward from the description using polynomials above. Therefore any element automorpmism of $\prod_UE/\prod_UF$ is determined by the image of $\alpha^*$, which on $U$-most indices must coincide with some conjugate of $\alpha$, and hence is induced by an automorphism of $E/F$. Conversely, any automorphism of $E/F$ induces an automorphism of the ultrapowers.

If $E/F$ is Galois but not finite, then $\prod_UE/\prod_UF$ need not a Galois extension, in fact it can fail to even be algebraic extension. This happens whenever $U$ is a nonprincipal ultrafilter on $\mathbb N$ (or more generally for any $U$ which is not countably complete). Indeed, in that case for any $n\in\mathbb N$ we can pick $\alpha_n\in E$ which has degree at least $n$ over $F$. Then $[(\alpha_n)]\in\prod_UE$ will not satisfy a polynomial equation of degree $\leq n$ over $\prod_UF$ for any $n$ by Łoś, so will be transcendental. You can still ask for the automorphism group of this extension, but I'm not sure it will have a nice explicit description. A natural guess would be that the automorphism group is $\prod_U\Gal(E/F)$, but I have a suspicion it will be larger than that.

If $E/F$ is infinite and $U$ is countably complete, then at the very least you can get that $\prod_UE/\prod_UF$ is algebraic and Galois. At least in the case of $E,F$ countable you should be able to get an isomorphism like in the first part of my answer, but I have no idea about general case.

$\newcommand{\Gal}{\operatorname{Gal}}$If $E/F$ is a finite Galois extension, then $\Gal(\prod_UE/\prod_UF)$ is canonically isomorphic to $\Gal(E/F)$. Indeed, by the primitive element theorem, $E=F(\alpha)$ for some $\alpha\in E$. This means every element of $E$ can be written as a polynomial in $\alpha$ with coefficients in $F$ of degree smaller than $d=\deg\alpha$.

Let $\alpha^*$ be the image of $\alpha$ in the ultrapower. Then we have $\prod_UE=(\prod_UF)(\alpha^*)$, as is straightforward from the description using polynomials above. Therefore any element automorpmism of $\prod_UE/\prod_UF$ is determined by the image of $\alpha^*$, which on $U$-most indices must coincide with some conjugate of $\alpha$, and hence is induced by an automorphism of $E/F$. Conversely, any automorphism of $E/F$ induces an automorphism of the ultrapowers.

If $E/F$ is Galois but not finite, then $\prod_UE/\prod_UF$ need not a Galois extension, in fact it can fail to even be algebraic extension. This happens whenever $U$ is a nonprincipal ultrafilter on $\mathbb N$ (or more generally for any $U$ which is not countably complete). Indeed, in that case for any $n\in\mathbb N$ we can pick $\alpha_n\in E$ which has degree at least $n$ over $F$. Then $[(\alpha_n)]\in\prod_UE$ will not satisfy a polynomial equation of degree $\leq n$ over $\prod_UF$ for any $n$ by Łoś, so will be transcendental. You can still ask for the automorphism group of this extension, but I'm not sure it will have a nice explicit description. A natural guess would be that the automorphism group is $\prod_U\Gal(E/F)$, but I have a suspicion it will be larger than that.

If $E/F$ is infinite and $U$ is countably complete, then at the very least you can get that $\prod_UE/\prod_UF$ is algebraic and Galois. At least in the case of $E,F$ countable you should be able to get an isomorphism like in the first part of my answer, but I have no idea about general case.

Edit: as Andreas Blass points out in a comment below, if $U$ is countably complete, then $\prod_U E\cong E$ if $E$ is countable. More generally, if $U$ is $\kappa$-complete, then $\prod_UE\cong E$ whenever $|E|<\kappa$. This in particular will hold if $U$ is countably complete and $|E|$ is below the first measurable.