Solving recurrence relation with square root
The answer cannot be $O(\log\log n)$. Already without applying any recursion we have the inequality $T(n) = T(\sqrt{n}) + n \ge n$. So the complexity cannot be smaller than $O(n)$. 



But now to your computation.  Setting $n=2^m$, we obtain as you did
$$ T(2^m) = T(\sqrt{2 ^ m}) + 2^m=T(2 ^ {\frac{m}{2}}) + 2^m.\tag{1}\label{eq1}$$ You defined $$S(m) = T(2^m).$$ Then equation $\eqref{eq1}$ should become the following equation,
which is different from $S(m)\,$$= S(\frac{m}{2})\,$$ + m$, the wrong equation in the question.

$$S(m) = S\left(\frac{m}{2}\right) + 2^m.$$

The equation above falls into the third case of the master theorem, therefore $S(m) \in \Theta(2^m)$. And from this follows $T(n) \in \Theta(n)$.