Q:

Why do we require quotient to be surjective?

Definition. Let $X$ and $Y$ be topological spaces; let $p : X \to Y$ a
  surjective map. The map $p$ is said to be a quotient map provided a
  subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.
Definition. If $X$ is a space and $A$ is a set and if $p : X\to A$ is a surjective map, then there exists exactly one topology $\tau$ on
  $A$ relative to which $p$ is a quotient map; it is called the quotient
topology induced by $p$.
Definition. Let X be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p : X  \to X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it. In the quotient topology induced by $p$, the space $X^ *$ is called a quotient space of $X$

It seems that to induce a quotient topology by using $p$, we don't need the property of $p$ being surjective. Why do we require a quotient map to be surjective? Are there some important applications of the surjection property? To prove a function is a quotient map, do we need to explicitly prove the function is surjective ? 
Does the following picture give some intuition? It's from Wikipedia. The next stage of this gif is the $S^2$ obtained by gluing the boundary (in blue) of the disk $D^2$ together to a single point.

A:

The quotient space has a universal property, namely if $f:X\rightarrow Z$ is a continuous function with the property that $(p(a)=p(b))\Rightarrow (f(a)=f(b))$, then $f$ factors through the quotient space.  In other words, there exists a unique continuous function $\widetilde{f}:Y\rightarrow Z$ so that $f=\widetilde{f}\circ p$.
$$
\begin{matrix}
X && \\
{\scriptsize p} \downarrow & \  \searrow^{f} & \\
Y & \underset{\bar f}{\longrightarrow} & Z 
\end{matrix}
$$
If you don't take the quotient map to be surjective, this property fails (the map fails to be unique).
From the definition you state, it seems that one could define a quotient even if map $p$ is not surjective, but this breaks a desired property (that might not have been mentioned in your studies yet).
If you consider a "quotient map" $q$ that has all the properties above, but isn't surjective, then you can write $q$ as a composition of a (true) quotient and an inclusion map.  In other words, the composition $X\stackrel{p}{\rightarrow} Y\stackrel{i}{\hookrightarrow}Z$ equals $q$.
$$
\begin{matrix}
X && \\
{\scriptsize p} \downarrow & \ \searrow^q & \\
Y & \underset{i}{\hookrightarrow} & Z
\end{matrix}
$$

