I cannot understand why this doesnt work for $v \neq 0$ and $u=0$? I think it may be to do with my lack of understanding of why $v=0$ and $u \neq 0$ works which I believe has come from the fact that if $f(x)=x^{1/2}$ then $\displaystyle f'(x)= \frac{1}{2x^{1/2}}$ so we must have that $x \neq 0$.
They did not say that it doesn't. The statement outside of parentheses is that the function is smooth everywhere except at $(u,v)=(0,0)$. This is easiest to see in polar coordinates: given a point other than the origin, represent it as $(r_0,\theta_0)$; then the functions are $$(r,\theta) \mapsto (\sqrt{ r},\theta/2),\qquad (r,\theta) \mapsto (\sqrt{r},\pi+ \theta/2)$$ which are well defined and smooth for $r>0$ and $|\theta-\theta_0|<\pi$.