In the proof of the quoted proposition, it is mentioned that $D^n \times I$ retracts onto $D^n \times \left\{0\right\} \cup \partial D^n \times I$ and an example is given in a figure with $n=2$, which is a radial projection from a point outside $D^2 \times I$.
Question 1: I can see that the radial projection maps $D^2 \times I$ onto $D^2 \times \left\{0\right\}$, but how does the part $\partial D^2 \times I$ arise from the radial projection? It seems that every point on the cylindrical boundary of $D^2 \times I$ (i.e. every point of $\partial D^n \times I$) will be sent by the radial projection on $D^2 \times \left\{0\right\}$. What am i missing here?
Next, it is claimed that $X^n \times I$ (that is the $n$-dimensional part of the cell complex $X$, whose $A$ is a subcomplex) is obtained from $X^n \times \left\{0\right\} \cup (X^{n-1} \cup A^n) \times I$ by attaching copies of $D^n \times I$ along $D^n \times \left\{0\right\} \cup \partial D^n \times I$.
Question 2: Why is the attachment of copies of $D^n \times I$ along $D^n \times \left\{0\right\} \cup \partial D^n \times I$ and not just along $\partial D^n \times I$? What is the role of $D^n \times \left\{0\right\}$?
Radial projection does not map $D^2 \times I$ onto $D^2 \times \{0\}$. Consider the side of the cylinder. All of these points are fixed by the projection.
You want to be able to preform a deformation retract. One can not do such a deformation retract to $\partial D^n \times I$ because you would have to be creating a hole in $D^n \times \{0\}$.