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.
Let $X$ be a topological space, $Y$ be a set and $f:X\to Y$ be a map of sets. We define the final topology $\mathscr{T}_f$ on $Y$ relative to $f$ to be the largest topology on $Y$ making $f$ continuous; this topology is exactely defined by $$\mathscr{T}_f=\{U\subseteq Y : f^{-1}(U) \ is \ open \ in \ X \}$$ Now consider an equivalence relation $\sim $ in $X$ and $\pi:X\to X/\sim$ the quotient map. The quotient topology in $X/\sim$ is defined to be the final topology on $X/\sim$ relative to $\pi$. Now if $X$ and $Y$ are topological spaces and $f:X\to Y$ is a surjection and a quotient map i.e, $U\subseteq Y$ is open iff $f^{-1}(U)\subseteq X$ is open. Then we define in $X$ the relation $\sim$ by: $x\sim x^\prime \Leftrightarrow f(x)=f(x^\prime)$. It is easy to see that $\sim$ is an equivalence relation and that $X/\sim$ endowed with the quotient topology is homeomorphic to $Y$.
Best Answer
As azif00 comments, the set $\sigma = \{\pi(V) \mid V \subset Q \text{ open}\}$ is not necessarily a topology on $\tilde Q$.
However, we can consider the topology $\tau(\sigma)$ having $\sigma$ as a subbase. Then $\pi : Q \to \tilde Q$ becomes an open map. But if we endow $\tilde Q$ with the quotient toplogy, $\pi$ is in general not an open map which shows that the two topologies are in general distinct. What is the reason for this fact?
The quotient topology $\tau_\pi$ is always a subset of $\sigma$. Its elements are the subsets $U \subset \tilde Q$ for which $\pi^{-1}(U)$ is open in $Q$, thus they have the form $U =\pi(\pi^{-1}(U))$ , i.e. are images of special open subsets of $Q$. But in general not every open $V \subset Q$ has the form $V = \pi^{-1}(U)$, thus in general $\tau_\pi \subsetneqq \sigma \subset \tau(\sigma)$.
Summarizing, $\tau(\sigma)$ is always finer than $\tau_\pi$ and in general even strictly finer.
Remark.
If $f : X \to Y$ is a function, a subset $S \subset X$ is called saturated if $S = f^{-1}(f(S))$. Note that $S \subset X$ is saturated if and only if it has form $S = f^{-1}(T)$ for some $T \subset Y$. For the only-if-part take $T = f(S)$. For the if-part observe that $S \subset f^{-1}(f(S))$ for all $S \subset X$ and $f(f^{-1}(T)) \subset T$ for all $T \subset Y$ so that $S \subset f^{-1}(f(S)) = f^{-1}(f(f^{-1}(T))) \subset f^{-1}(T) = S$ which implies $S = f^{-1}(f(S))$.
Thus the open subsets of $Q$ having the form $\pi^{-1}(U)$ for some $U \subset \tilde Q$ are precisely the open saturated subset of $Q$. Therefore $U \in \tau_\pi$ if and only if it is the image of an open saturated subset of $Q$. In fact, $U \in \tau_\pi$ means that $\pi^{-1}(U)$ is open in $X$. As we have seen above, $\pi^{-1}(U)$ is saturated and $\pi(\pi^{-1}(U)) = U$ since $\pi$ is surjective; this proves the only-if-part. For the if-part let $U = \pi(V)$ with an open saturated $V \subset Q$. Then $\pi^{-1}(U) = \pi^{-1}(\pi(V)) = V$ is open in $Q$, thus $U \in \tau_\pi$.