Transition Function on the 2 - sphere In this problem, we consider $S^2$ with the coordinate charts $(U, \phi)$, and $(U', \phi')$, where:
$$U = \{(x,y,z) \in S^2 : z < 0\} \quad \phi(x,y,z) = (x,y), \ \mathrm{and}$$
$$U'\ = \{(x,y,z) \in S^2 : x > 0\} \quad \phi'(x,y,z) = (y,z)$$
My question concerns explicitly writing out the sets $\phi(U \cap U')$, and $\phi'(U \cap U')$. My attempt for writing out $\phi(U \cap U')$ is: 
$$\phi(U \cap U') = \{(x,y) \in \mathbb{R^2} : x = \sqrt{1 - y^2 - z^2}\}$$
This incorporates the fact that $x > 0$ and $(x,y,z) \in S^2$. However, since $\phi$ drops the $z$ term, it feels inappropriate the include a $z$ in the definition of $\phi(U \cap U')$, and this makes it difficult to invert $\phi$ when computing the transition function. I am confused as to how to write the set explicitly that makes it simple to compute the transition function.
 A: I think that the confusion lies in that you are assuming the image of $\phi(U\cap U^{\prime})$ has to line in $S^{2}$, when really it lies in $\mathbb{R}^{2}$. To elaborate, I'll write $U_{x,+} = \{(x,y,z)\in S^{2} : x > 0 \}$ and $\phi_{x,+}(x,y,z) = (y,z)$ for its coordinate chart. But since $\phi_{x,+}:U_{x,+} \rightarrow \phi_{x,+}(U_{x,+}) = V_{x,+}$, say its inverse $\phi_{x,+}^{-1}:V_{x,+} \rightarrow U_{x,+}$ has to have its image lie within $U_{x,+} \subset S^{2}$, so we take 
$$
\phi_{x,+}^{-1}(y,z) = \bigg(\sqrt{1 - y^{2} - z^{2}}, y, z\bigg) \in U_{x,+} \subset S^{2}.  
$$
Very similar coordinate functions for the other coordinate neighbourhoods have similar properties. As to your question, I would say that the information regarding $z$ is encoded more subtly in the form of the inverse coordinate functions (note that whilst these choices of coordinate functions are usually many-to-one if the domain was $\mathbb{R}^{3}$, the fact that it is $S^{2}$ then forces them to be injective.). 
EDIT - Further elaboration: I'm basing this on an example in Do Carmo's 'Riemannian Geometry', on page 21.
The coordinate charts can be seen as orthogonal projections from $S^{2}$ onto some subset of $\mathbb{R}^{2}$, e.g.
$$
\phi_{x,+}(x,y,z) = (0,y,z) = (y,z) \in \mathbb{R}^{2}.
$$
In the ambient $\mathbb{R}^{3}$ however, say after the inclusion $(y,z) \mapsto (y,z,u)$ there would indeed be two points in $\mathbb{R}^{3}$ such that $y^{2} + z^{2} + u^{2} = 1$, and indeed making a choice of sign will send us back to a unique point in $S^{2}$, so the inverse chart function would be e.g.
$$
\phi_{z,-}^{-1}(u,v) = (u,v,-\sqrt{1-u^{2}-v^{2}}) \in U_{z,-}.
$$
Combining these, say on $\phi_{x,+}(U_{x,+}\cap U_{z,-})\subset \mathbb{R}^{2}$, then:
$$
\phi_{z,-}\circ \phi_{x,+}^{-1}(u,v) = \phi_{z,-}(\sqrt{1 - u^{2} - v^{2}},u,v) = (\sqrt{1 - u^{2} - v^{2}},u)\in \phi_{z,-}(U_{x,+}\cap U_{z,-})
$$
