This is a follow-up question to Homotopy on $S^{n-1}$ extends to homotopy on $D^n$ .
I doubt that the retraction $r_n:D^n\times I\to S^{n-1}\times I\cup D^n\times\{0\}$ will help you in the sense you expect - but of course we can use it. There exists a homeomorphism
$h : D^n \times I \to D^{n+1}$ such that $h(D^n \times \{0,1\} \cup S^{n-1} \times I) = S^n$. Note that $D^n \times \{0,1\} \cup S^{n-1} \times I$ is the boundary of $D^n \times I$.
The desired retraction is
$$R_n : D^n \times I \times I \stackrel{h \times id_I}{\to} D^{n+1} \times I \stackrel{r_{n+1}}{\to} S^n \times I \cup D^{n+1} \times \{0\} \stackrel{h^{-1} \times id_I}{\to} \\ h^{-1}(S^n) \times I \cup h^{-1}(D^{n+1}) \times\{0\} = \\ (D^n \times \{0,1\} \cup S^{n-1} \times I) \times I \cup D^n \times I \times \{0\} = \\ (S^{n-1}\times I\times I)\cup (D^n\times\{0,1\}\times I)\cup (D^n\times I\times \{0\}) .$$
Let us come to the construction of $h$. Of course this is equivalent to the construction of a homeomorphism $H : D^n \times [-1,1] \to D^{n+1}$ such that $H(D^n \times \{-1,1\} \cup S^{n-1} \times [-1,1]) = S^n$.
Let $\lVert - \rVert^{(k)}$ denote the standard Euclidean norm on $\mathbb R^k$. On $\mathbb R^{n+1}$ define $\lVert (x_1,\ldots,x_{n+1}) \rVert = \max(\lVert (x_1,\ldots,x_n) \rVert^{(n)},\lvert x_{n+1} \rvert)$. This a norm on $\mathbb R^{n+1}$ which is equivalent to $\lVert - \rVert^{(n+1)}$. The closed unit ball with respect to $\lVert - \rVert$ is $D^n \times [-1,1]$. Define
$$\phi : \mathbb R^{n+1} \to \mathbb R^{n+1}, \phi(x) = \begin{cases} \dfrac{\lVert x \rVert}{\lVert x \rVert^{(n+1)}} x & x \ne 0 \\ 0 & x = 0 \end{cases}$$
This a homeomorphism such that $\phi(D^n \times I) = D^{n+1}$. See my answer to Homeomorphism between $k$-simplex and a product of $k$ unit intervals .
Restriction to $D^n \times I$ gives the desired $H$.
