Same construction works with any $g:[a,b]\rightarrow \mathbb{R}$ with $b=a^2$, there's nothing special about $2,4$ right?

@Yanko Yes, there's nothing special about $[2,4)$, it can be any interval $[a,a^2)$ with $a>1$

I think you need the additional condition that g(4)=g(2)+1, else we might not have continuity at the places where you are gluing together the pieces of the piecewise definition. Otherwise, good answer!

@Angelo I doubt there's a simpler way to characterize the solutions. For $g(x) = \sin(x)$, calculating $f(5/4)$ takes some lengthy calculation, but after a little coding, I was able to calculate not only $f(5/4)$ but to plot it really fast in the interval $(1,10)$

The definition for $x\in (1,2)$ is given in the answer. It's $f(x^2)-1$

@FranklinPezzutiDyer This gives ALL the solutions, not only the continuous solutions.

If $x \in (1, \sqrt 2)$ then $x\in (1,2)$, so you apply the corresponding definition: $f(x^2)-1$

The definition is recursive, which I also mentioned in the answer. You may need to apply the definition several times before reaching the interval $[2,4)$. If a programming language was able to understand my definition, surely you also can.

When see the definition of the Fibonacci numbers $F_{n+2} =F_{n+1} + F_n$ with $F_0=0$ and $F_1=1$, do you also demand a better definition because the definition don't give explicitly all the values of the sequence?

Fibonacci definition is good. You should use the index $n$ too in order to improve your recursive definition.

You should write something like $f(x)=\begin{cases}g(x)&\text{if }\;2\leqslant x<4\[5pt]f\!\left(\sqrt[2^n]x\right)+n&\text{if }\;2^{2^n}\leqslant x<4^{2^n}\[5pt]f\!\left(x^{2^n}\right)-n&\text{if }\;\sqrt[2^n]2\leqslant x<\sqrt[2^n]4\end{cases}\;\;$ where $\,n\in\Bbb N\,.\;$

All the continuous functions $\,f:(1,\infty)\to\mathbb{R}\,$ that satisfy $\,f(x^2)=f(x)+1\,$ are the following ones: $$f(x)=\begin{cases}g(x)&\text{if }\;2\leqslant x<4\[5pt]f\!\left(\sqrt[2^n]x\right)+n&\text{if }\;2^{2^n}\leqslant x<4^{2^n}\[5pt]f\!\left(x^{2^n}\right)-n&\text{if }\;\sqrt[2^n]2\leqslant x<\sqrt[2^n]4\end{cases}$$ where $\,n\in\Bbb N\,$ and $\,g:[2,4)\to\Bbb R\,$ is any continuous function such that $\,\lim\limits_{x\to4^-}g(x)=g(2)+1\,.\;$

@FranklinPezzutiDyer The original question didn't ask for continuous solutions, so I do answered the OP's question.

@Angelo I should write $f$ like that? Why? You're giving an explicit solution, I prefer the recursive solution. And as a matter of taste I prefer my way of writing it.

@jjagmath, your way to write the recursive solution is not correct, differently from fibonacci recursive definition which is good. From your definition it is not clear how you get $f(x^2)$ if $1<x<2$. You should explain it in a better way.

@Angelo Why do you insist it's not correct. Let say you want to calculate $f$ when $x = 5/4 \in (1,2)$. The definition of $f$ is $f(5/4) = f(25/16)-1$. Now we need to calculate $f(25/16)$, since $25/16 \in (1,2)$ the same part of the definition applies, so $f(25/16) = f(625/256)-1$. And now $625/256 \in [2,4)$, so $f(625/256) = g(625/256)$, giving $f(25/16) = g(625/256)-1$ and finally $f(5/4) = g(625/256)-2$. I followed the definition I gave and was able to calculate $f$. And the same will happen with all the numbers in $(1,2)$, only with a different number of recursions.

@Angelo You said "From your definition it is not clear how you get $f(x^2)$ if $1<x<2$. You should explain it in a better way" . My answer was only the definition of the function. I didn't prove anything because it wasn't my intention. To give a thorough answer one will need to prove that the recursion ends, prove that the function is indeed a solution and then prove that any solution is obtained that way. But that doesn't make the definition of $f$ incorrect.

Is it possible to continue the solution? I have no clue what to do next.