First, one constructs a cylinder over the space $X$, that is, $X\times I$ (here $I=[0,1]$). Then one maps the top of the cylinder into $Y$. You can think on it as gluing the top $X\times \{1\}$ with the image $f(X)$.
Thanks to @Danu for the reference. I will give an outline of the proof in Hatcher:
Claim: A map $f : X \to Y$ induces a homomorphism $f_*: H_n(X) \to H_n(Y)$.
Proof: First, define an induced homomorphism $f_\# : C_n(X) \to C_n(Y)$ by $(\sigma : \Delta^n \to X) \mapsto (f \circ \sigma : \Delta^n \to Y)$.
Secondly, we claim that $f_* : H_n(X) \to H_n(Y) : [a] \to [f_\#(a)]$ is a homomorphism. We need that $f_\#(a) \in \ker \partial_n$ whenever $a \in \ker \partial_n$ (so the map makes sense) and $f_\#(b) \in \text{Im }\partial_n$ whenever $b \in \text{Im }\partial_n$ (so the map is well defined, independent of the choice of representative $a$). ($\partial$ is the differential function $\partial_n : C_n(X) \to C_{n-1}(X)$ or $\partial_n : C_n(Y) \to C_{n-1}(Y)$.)
We can check $f_\# \partial = \partial f_\#$ by explicit computation and this gives us what we need.
We have two easy results:
1) $(f \circ g)_* = f_* \circ g_*$ and
2) id$_{*X} = \text{id}_{H_n(X)}$ (the induced map of the identity on $X$ is the identity on $H_n(X)$).
Proof of 1): We have $(f \circ g)_\# = f_\# \circ g_\#$ (by associativity of $\circ$) and therefore $(f \circ g)_* = f_* \circ g_*$.
Proof of 2): By definition id$_{\#X}$ is id$_{C_n(X)}$ and so id$_{*X} = \text{id}_{H_n(X)}$.
Theorem: If two maps $f,g : X \to Y$ are homotopic, then they induce the same homomorphism $f_* = g_* : H_n(X) \to H_n(Y)$.
The proof of this is the meat of the problem and is given in Hatcher (p111-113).
Corollary: The maps $f_* : H_n(X) \to H_n(Y)$ induced by a homotopy equivalence $f: X \to Y$ are isomorphisms for all $n$.
Proof: If $f$ is a homotopy equivalence then there is a map $g$ such that $f \circ g$ is homotopy equivalent to id$_Y$ and $g \circ f$ is homotopy equivalent to id$_X$.
Then, by the theorem, $(g \circ f)_* = id_{*X}$ and $(f \circ g)_* = id_{*Y}$. Using the two easy results, we get $g_* \circ f_* = id_{H_n(X)}$ and $f_* \circ g_* = id_{H_n(Y)}$.
Note that a deformation retraction is a special case of a homotopy equivalence so I've now also shown that homologies are invariant under deformation retraction.
Best Answer
I'm not sure how to interpret the first question, but there are a few answers.
1) It is a quotient space because it is a quotient. You define the mapping cylinder of $f\colon X\to Y$ as the quotient of the space $X\times[0,1]\amalg Y$ by the relation $(x,1)\sim f(x)$.
2) A surjective map $q\colon A\to B$ is a quotient map if and only if "a set $U$ is open in $B$ if and only if $q^{-1}(U)$ is open in $A$." In other words, $B$ has the quotient topology. In this case, the quotient map is $X\times[0,1]\amalg Y\to(X\times[0,1]\amalg Y)/\sim$.
3) In another sense, to "be a quotient space" means nothing. Any space $X$ is a quotient space of itself with the trivial relation.
As to the second question, there is no significance to the circled area. Hatcher just tried to draw something more interesting than a circle.
The point of mapping cylinders is really so that, for any map $f\colon X\to Y$, you can replace $f$ with the inclusion of $X$ into the mapping cylinder $M_f$. Here, "inclusion" means the map $i\colon X\to M_f$ sending $x$ to $(x,0)\in X\times[0,1]$. Then via a straight-line deformation, $M_f\simeq Y$. The point is that, when you want to compute the homology of one of the spaces $X$ or $Y$, you can now think of $X$ (i.e. $X\times\{0\}$) as being a subspace of $Y$ (i.e. $M_f$), and then look at the long exact sequence of the pair $(M_f,X)$.