In Allen Hatcher Algebraic topology, chapter zero, I had the following:
My questions are:
1- Why the mapping cylinder is the quotient space, what makes it specifically a quotient space?
2- Why there is the part I circled in red in the picture in the mapping cylinder, what is its importance?




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 $(M_f,X)$.