You're right, Munkres has given a very confusing picture. (Also, the term saturated is hardly standard in this context --- I've never seen in used in other topology books.)
For (1): you are correct about $V$. For $U$, your answer is a possible correct answer, but it would also be OK not to include the outer boundary; in that case, the corresponding open set on the sphere would look like a punctured open disk, i.e., missing a point inside.
For (2): the idea is that we can take the upper hemisphere and stretch it over the entire sphere, collapsing the equator to a single point (the south pole). To describe this formally, let the disk on the left have radius $\pi$. Use polar coordinates, so a point in the disk is $(r,\theta)$. On the sphere, use coordinates $(\alpha,\beta)$ where $\alpha$ is the angle of latitude, but measured from the north pole (so $\alpha=0$ at the north pole, $\alpha=\pi$ (i.e., $180^\circ$) at the south pole), and $\beta$ is the angle of longitude.
To be quite explicit about the definition of $\alpha$: In the diagram latitude, the latitude is the angle ϕ. However, you'll note that ϕ is measured up from the equator. That's standard, of course. I want an angle like ϕ, but measured down from the north pole. So in the northern hemisphere, α=π/2−ϕ (using radians, π/2=90∘), and in the southern hemisphere, α=ϕ+π/2.
Now let $(r,\theta)$ map to the point with $\alpha=r, \beta=\theta$. When $r=\pi$, we have the entire circumference mapping to the south pole.
Imagine that you place a disk on top of a sphere so that the center of the disk is at the north pole. Then spread the disk over the surface of the sphere, and glue the boundary of the disk at the south pole. This is one way to see the homeomorphism that the book is talking about.
To construct such a homeomorphism explicitly, we can use polar coordinates $(r, \theta)$ in $\mathbb R^2$ and cylindrical coordinates $(r, \theta, z)$ in $\mathbb R^3$. Consider the continuous map $f : D^2 \to S^2$ given by
$$
(r, \theta) \mapsto \left(\sqrt{1 - (1 - 2r)^2}, \theta, 1 - 2r\right).
$$
Since the boundary of $D^2$ (corresponding to $r = 1$) is mapped to a single point (the south pole), it follows that $f$ factors through a map $\overline f : X^* \to S^2$, where $X^*$ is $D^2$ with the boundary identified to a single point. By the familiar properties of quotient maps, we see that $\overline f$ is a continuous bijection from a compact space to a Hausdorff space. We conclude that it is the desired homeomorphism.
Best Answer
HINT: Design your homeomorphism $h$ so that it takes:
You may find it helpful to think in terms of cylindrical coordinates.