Is the square root of a monotonic function whose all derivatives vanish smooth?

Extend the domain of the function $g$ to $\mathbb R$ by letting $g(x):=0$ for real $x<0$. The resulting function, which we shall still denote by $g$, is $C^\infty$ on $\mathbb R$.

Theorem 3.5 on page 144 implies that a nonnegative $C^4$ function $f$ on $\mathbb R$ has a $C^2$ square root if for any minimum $x_0$ of $f$ we have $f(x_0)=0$.

This latter condition obviously holds for our function $g$ in place of $f$ -- because $g$ is strictly increasing on $[0,\infty)$ and thus has no minima in $(0,\infty)$, and $g=0$ on $(-\infty,0]$.

Therefore, we can conclude that $\sqrt g$ is $C^2$ on $\mathbb R$, even without assuming that $g''>0$ in a neighborhood of zero.

Yet, this conclusion falls short of your main goal, to show that $\sqrt g$ is $C^\infty$. Looking at the proof of the mentioned Theorem 3.5, this task may be too big for a usual MO answer and may require a full-blown paper.

The answer is yes, by the results of

Bony, Jean-Michel; Colombini, Ferruccio; Pernazza, Ludovico, On square roots of class (C^m) of nonnegative functions of one variable, Ann. Sc. Norm. Super. Pisa, Cl. Sci. (5) 9, No. 3, 635-644 (2010). ZBL1207.26004.

Here is the math review:

Clearly the condition is fulfilled in the OP (for any $m$).