Define the reduced mapping cylinder as follows:
Suppose we start with continuous map $f : (X,y_0) \rightarrow (Y,y_0)$ and define the reduced mapping cylinder as follows.
$$M_f: = \mathbb{I} \times X \amalg Y/ \sim \quad , (x,1) \sim f(x), \quad \textit{and } \quad (x_0,t) \sim y_0 $$.
I constructed a retraction $r : (M_f,\star) \rightarrow (Y,y_0)$ just from universal property of quotients.
I constructed the following homotopy $\bar{H} : (M_f,\star) \times \mathbb{I} \rightarrow (M_f,\star)$ defined as
$\bar{H}([y,t]) = [y]$, $\bar{H}([x,s],t) = [x,t + (1 - s)s] if [x] \neq [x_0]$ and $\bar{H}([x_0,s],t) = [y_0] if [x] = [x_0]$. I got this map from geometrical reasoning i.e we are sliding the point [x,s] into its f(x) point. We can check that $\bar{H}$ is well defined that is easy.
I have two main issues which I would like to understand:
First issue: Why is $\bar{H}$ continuous ? I believe it should be for geometrical intuition.
Second issue: We have $\bar{H}(-,0) = id$ but why is $\bar{H}(-,1) = t \circ r$ where $t : Y \rightarrow M_f$ where it is defined as $y \mapsto [y]$ ?
There is no geometric reason for the continuity of $\overline{H}$. Instead, consider the quotient map $$p:(I\times X)\coprod Y\rightarrow M_f.$$ Since $I$ is locally compact, $$\text{id}\times p :I\times (I\times X)\coprod Y\rightarrow I\times M_f$$ is also a quotient map. Define the map $$H:I\times (I\times X)\coprod Y\rightarrow M_f$$ by $ H(s,t,x)= [(s+(1-s)t,x)]$ and $H(y)=[y]$. Then $H$ is continuous and $H = \overline{H}\circ \text{id}\times p$. Since $\text{id}\times p$ is a quotient map, $\overline{H}$ is also continuous.
For your second question, I assume that you defined $r([t,x])=f(x)$ and $r([y])=y$. Then by definition $$(t\circ r)([t,x])=[f(x)]=[1,x]=\overline{H}([1,t,x])$$ and $$(t\circ r)([y])=[y]=\overline{H}([1,y]).$$