For example, $\ln(x^y) = \ln(x) \times y$, so $\ln$, so to speak, "distributes" into the first argument of exponentiation, while turning it into multiplication.
I wonder if there is a function that "distributes" into the second argument of exponentiation, but I have no idea how to even approach the search.
P.S. There are of course some trivial solutions, like for example $f(x) = x$. Of interest though are those that do not involve computing $x^y$.
$$f(x^y) = g(x, f(y))$$
Substitute $y=f^{-1}(v)$. Then $$f(x^{f^{-1}(v)})=g(x,v)$$ and so we have a general formula for $g$.
Therefore the perhaps surprising answer is that all functions $f$ can be said to satisfy an equation of this form.
For any function $f$, simply define $g$ by $$g(u,v)=f(u^{f^{-1}(v)}).$$
For example, if $f$ is the doubling function then $$g(u,v)=2(u^{\frac{v}{2}}).$$
You can use this to explore what happens for different $f$. The real issue is perhaps whether one can find a function $f$ which gives a "nice" function $g$ as happened in your opening example.