In my experience the usual context in which saturated sets appear in topology is the one metioned by Stefan H. in the comments: you have a map $f:X\to Y$, and you say that a set $A\subseteq X$ is saturated with respect to $f$ iff $A=f^{-1}\big[f[A]\big]$. More generally, if $\mathscr{P}$ is a partition of $X$, a set $A\subseteq X$ is saturated with respect to $\mathscr{P}$ iff $A=\bigcup\{P\in\mathscr{P}:P\cap A\ne\varnothing\}$. (This really is a generalization: in the case of the map $f$, the associated partition of $X$ is $\{f^{-1}[\{y\}]:y\in Y\}$, the set of fibres of the map $f$.)
If $\mathscr{P}$ is a partition of $X$ and $A$ is an arbitrary subset of $X$, there are two saturated sets naturally associated with $A$. One, which we might call the outer saturation of $A$, is $$\bigcup\{P\in\mathscr{P}:P\cap A\ne\varnothing\}\;;$$ if $\mathscr{P}$ is generated as above by a map $f:X\to Y$, the outer saturation of $A$ is $A=f^{-1}\big[f[A]\big]$. The other, which we might call the inner saturation of $A$, is $$\bigcup\{P\in\mathscr{P}:P\subseteq A\}\;;$$ it’s the complement of the outer saturation of $X\setminus A$, so if $\mathscr{P}$ is generated as above by a map $f:X\to Y$, the inner saturation of $A$ is $X\setminus f^{-1}\big[f[X\setminus A]\big]$.
My best guess without having seen the actual context is that by the saturation of a set $A$ they mean what I’ve called here the outer saturation of $A$.
Note: The terms outer saturation and inner saturation are not standard, so far as I know; I’m using them here for purposes of exposition.
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.
Best Answer
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.