The definition of a quotient map is not very enlightening, in my opinion. The intuition behind $X/\sim$ is "crushing the equivalence classes to points" inside of $X$. This is best seen through some examples:
The interval $[0,1]$ with the relation $0\sim 1$ gives the quotient $[0,1]/\{0,1\} \cong S^1$, the circle.
More generally, $D^n/\partial D^n \cong S^n$, where $D^n$ is the closed $n$-disk and $S^n$ is the $n$-sphere.
But how does this relate to the technical definition? First, $X/\sim$ as a set should not distinguish between two points in the same equivalence class. Thus, it's natural to take $X/\sim$ as the set of equivalence classes.
As for the topology, for the quotient $f: X\to X/\sim$ to be continuous, we must require that for any open set $U\subset X/\sim$, we have $f^{-1}(U)$ is open in $X$. But we want the topology of $X$ to entirely determine the topology of its quotient, so it's natural to define the open sets of $X/\sim$ to be precisely the subsets with open preimage in $X$. Thus, we recover the definition.
Edit: Now, what do we mean by "passing to the quotient"? This is relatively easy to understand if we think of the quotient space $X/\sim$ as crushing the equivalence classes to points: If we have a continuous map $f: X \to Y$ (for some arbitrary space $Y$) that is constant on some equivalence class $S$ (say $f$ maps points in $S$ to the point $y\in Y$), then we can think of $f$ mapping the entire equivalence class $S$ to $y$. Hence, we can see $f$ as a map on the quotient $f:X/\sim \to Y$ where $f$ maps the equivalence class $S$ to $y$. If $f$ is constant on each equivalence class, this gives a well-defined mapping, so $f$ "passes" or "descends" to the quotient.
Your question is too vague to give a precise answer to. After all, I don't understand what your Y is or why you think it was assembled by gluing or why it still sits in $\Bbb R^3$. Until you can tell me what gluing is I can't tell you why that's what quotients are.
Still, here is the result that unifies all of your three examples. I leave the proof as an exercise to you.
If $X$ is a compact space, and $Y$ a Hausdorff space, and $f: X \to Y$ is a continuous surjection, then $f$ descends to a homeomorphism $X/\sim \to Y$, where $x \sim y$ if $f(x) = f(y)$.
Whatever your space Y is, when you tell me that you are imagining gluing points of X together, you are presumably telling me a map from X to Y. That's what the above is. Then the result says that Y is precisely the quotient space obtained by identifying the points in X that f does.
In your first example, f traces out the circle counterclockwise. The second example is similar. In your third example, f stretches a rubber disc over the 2-sphere with the entire boundary of that disc getting mapped to the bottom point.
Best Answer
Perhaps the simplest interesting example is the quotient of $[0,1]$ obtained from the equivalence relation $E$ whose equivalence classes are the singletons $\{x\}$ for $0<x<1$ and the doubleton $\{0,1\}$. This identifies the endpoints $0$ and $1$ to a single point, and the quotient space is homeomorphic to $S^1$, the circle. Taking $S^1$ to be specifically the unit circle in the plane, one homeomorphism is the map
$$h:[0,1]/E\to S^1:p\mapsto\begin{cases} \langle\cos2\pi x,\sin2\pi x\rangle,&\text{if }p=\{x\}\\\\ \langle 1,0\rangle,&\text{if }p=\{0,1\}\;. \end{cases}$$
The quotient topology is exactly the one that makes the resulting space ‘look like’ the original one with the identified points glued together.
(This is really just the beginnings of an answer, because I’m not sure exactly what you want to know.)