Consider two parameters $\sigma(u,v)$ and $\sigma'(u',v')$ of $x^2+y^2+z^2=1$, where $x,y,z>0$. Given $\sigma(u,v) = (u,\sqrt{1-u^2-v^2},v)$ and $\sigma'(u',v') = \left(\sqrt{1-u'^2-v'^2},u',v'\right)$, where $u,v,u',v' \in U = \left\{(x,y):x>0,y>0,x^2+y^2<1\right\}$.
Find $\phi = \sigma'^{-1} \circ \sigma$.
HINT
The basic idea is that $\sigma$ picks some $(a,b)$ from the unit circle's first quarter (which we will call first quarter) and maps to the upper unit hemisphere as $(a, \cdot, b)$ and $\sigma'$ maps $(a,b)$ to the right unit hemisphere as $(\cdot,a,b)$.
So the point $(a,b)$ from the first quarter gets mapped to $\left(a, \sqrt{1-a^2-b^2},b\right)$ under $\sigma$ and then to find the image under $\sigma'^{-1}$, you answer the question which $c,d$ do we have to apply $\sigma'$ to in order to get $\left(a, \sqrt{1-a^2-b^2},b\right)$?
Can you take it from here? Please feel free to post further questions in comments if you need more guidance.