The quotient space $Y = X / \sim$ as a set is just the set of equivalence classes of $X$ under $\sim$, so the set $\{ [x]: x \in \mathbb{R} \} $ in your case.
The equivalence class of a number $x$ is just (in your case) the set $\{ x+n : n \in \mathbb{Z} \}$. Now we need a topology. The standard topology that we take on $Y$ is all subsets $O$ of $Y$ (where points in $Y$ are "really" subsets of $X$, the equivalence classes) such that $q^{-1}[O]$ is open in the topology of $X = \mathbb{R}$. Here $q$ is the map that sends $x$ to its class $[x]$ in $Y$, the so-called quotient map. This is called the quotient topology on $Y$, and as you see it assumes you have a topology on $X$ already, and we give $Y$ the largest topology possible to still have $q$ continuous. (The smallest one would always be the indiscrete topology, which is not very interesting, hence the other "natural" choice.)
Now, if we have a function $f$ from $Y$, the quotient space in the quotient topology, to any space $Z$, then $f$ is continuous iff $f \circ q$ is continuous as a function from $X$ to $Z$: one way is clear, as the composition of continuous maps is continuous, and for the other side, if $O$ is open in $Z$, by definition $f^{-1}[O]$ is open in $Y$ iff $q^{-1}[f^{-1}[O]]$ is open in $X$, and this set equals $(f \circ q)^{-1}[O]$ which is open, as by assumption $f \circ q$ is continuous.
Now, consider the map $f$ that sends the class $[x]$ to the point $e^{2\pi ix}$ in $\mathbb{S}^1$, the unit circle.
This is well-defined: if $x'$ were another representative of $[x]$, then $x \sim x'$ and thus $x - x'$ is an integer and so $f(x') = f(x)$.
It is continuous, as $f \circ q$ is just the regular map sending $x$ to $e^{2\pi ix}$, and this is even differentiable etc.
It is clearly surjective and injective because the only way $[x]$ and $[y]$ will have the same value is when $2 \pi ix - 2 \pi i y$ is an integer multiple of $2 \pi i$, which happens iff $x - y$ is an integer.
One can also check that $q[X] = q[[0,1]]$ and by continuity of $q$ we have that $Y$ is compact. This makes (with $\mathbb{S}^1$ Hausdorff) the map $f$ a homeomorphism, by standard theorems.
We could also have achieved this as the quotient of $[0,1]$ under the equivalence relation that has exactly one non-trivial class, namely $\{0,1\}$. This is more intuitive, as we then glue together (consider as one point) just the points $0$ and $1$, and this geometrically gives a circle. In your example (which is a nice so-called covering map, and a group homomorphism as well) we glue a lot more points together, but all classes are now similar: just shifted versions of a point by an integer. We sort of wrap the interval $[0,1)$ infinitely many times over itself.
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
Consider the square $[0,1]\times [0,1]$, when you say identify $(0,t)\sim(1,t)$ it means, rolled up a piece of paper of size $[0,1]\times [0,1]$ in a way that, upper edge overlap with lower edge, and this gives you a cylinder with horizontal axis. Now we also assume that, $(t,0)\sim (t,1)$ on the square $[0,1]\times [0,1]$. But the points $(t,0),(s,1)$ with $0\leq t,s\leq 1$ now are on the two circular edges of cylinder. So to identify $(t,0)\sim(t,1)$ in square is now same as identifying the point $c_{(t,0)}$ on the left hand side of cylinder to the $c_{(t,1)}$ on the right hand side of the cylinder. So you have to bent the cylinder such that, $c_{(t,0)}$ coincide with $c_{(t,1)}$ i.e. join the left side of the cylinder to the right side of the cylinder without any perturbation or twist. This gives you torus.
Now, the edges of the square $[0,1]\times [0,1]$ after identification gives you a shape of figure-eight or $\Bbb S^1\lor \Bbb S^1$ on the torus $T$. While the interior points of $[0,1]\times [0,1]$ will associate to the points of $T\backslash (\Bbb S^1\lor\Bbb S^1)$ in one-to-one and onto fashion. Here $\Bbb S^1\lor\Bbb S^1$ means the joining of two circles exactly at one point.
While constructing torus from $[0,1]\times[0,1]$ if you consider the edges of this square, then as a first step of the identification $(t,0)\sim (t,1)$ gives you a line segment with two circles hanging on two ends of the segments. Next, as a second step of the identification $c_{(0,t)}\sim c_{(1,t)}$ ends up with identifying two ends of the previous line segment as well as identifying the two circles which are hanging two ends of the line segment, i.e., a figure with a shape like two circles joined or wedge exactly at one point, and this wedge point is nothing but the equivalence class $\{(0,0),(0,1),(1,0),(1,1)\}\in T$.