$\color{olive}{T(n) = 2T(\sqrt{n}) + log(n)}$
Let $\color{navy}{n = 2^m}$, and putting this in above recurrence relation.
$T(2^m) = 2T(2^{\frac{m}{2}}) + m$
Let, $S(m) = T(2^m)$, then, $S(\frac{m}{2}) = T(2^\frac{m}{2})$
$S(m) = 2S(\frac{m}{2}) + m$
This is similar to Merge-Sort (or) using master's theorem, it can be said that $S(m) = O(mlog_{2}(m))$
Now. replacing $m$ by $log_{2}(n)$, $\color{red}{T(n) = O(log_{2}(n)log_{2}log_{2}(n))}$