Let $f(x)$ be the half-iterate of $ 2 sinh(x).$
Im looking for an asymptotic to $f( 2 f^{[-1]}(x))$ for Large $x>0$. $^{[*]}$ means iteration here thus $^{[-1]}$ means functional inverse.
For the $y$ th iteration of $ 2 sinh(x) $ we use the fixpoint at $0$ And use the so-called koenigs function. Also we use analytic continuation.
See
https://en.m.wikipedia.org/wiki/Koenigs_function
Let O be big-oh notation. I think the approximation below makes sense
$$ f( 2 f^{[-1]} (x) ) = O ( x * 2sinh^{[A]}(x) ) $$
For some real $A>0$ ??
What is the best fit value for $A$ ?
$f(f(x))=2sh(x)$ and let put $A=f(2f^{-1}(x))$
If $y=f^{-1}(x)$ then $f(x)=2sh(y)$ and $f^{-1}(x)=argsh(\frac{f(x)}{2})$
Thus $f(A)=2sh(2argsh(\frac{f(x)}{2}))$ and reporting in $f^{-1}$ formula
$A=argsh(f(2sh(2argsh(\frac{f(x)}{2}))/2)$
So if we suppose $\lim \limits_{x\to+\infty} f(x)=+\infty$ we can use $argsh(\frac{u}{2})\sim\ln(u)$ and $2sh(u)\sim e^u$
$A\sim \ln(f(2sh(2\ln(f(x))))\sim \ln(f(e^{2\ln(f(x)}))\sim \ln(f(f(x)^2)$
All this is not very rigourous, but I suppose that if you know some equivalent for $f$ in $+\infty$ then you can conclude easily. I'm not familiar enough with half-iterate of $sh$ to go further.