If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there
but also the derivatives must be the same, so that $1=\cos z$ there.
Then $\sin^2z = 1-\cos^2 z =0$ and therefore $z=\sin z=0$, so that the only double zero is at the origin. If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{h(z)}$.  Since $\sin z$  grows at most exponentially, the growth of $Re\; h(z)$ cannot be quadratic or higher in $|z|$.  Hence $h$ must be a linear polynomial.  But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.]  See [Borel-Caratheodory](https://en.wikipedia.org/wiki/Borel%E2%80%93Carath%C3%A9odory_theorem).

Hence the square root cannot be an entire function.  The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.