Let $i_p$ be the constant map at $p$ being the origin of the path $f$. Prove that: $f * f^{-1} \simeq i_p \text{ rel } \dot I$.
The proof says to subdivide $I \times I$ and let the homotopy be:
$$H(s,t) = \begin{cases} f(2s(1-t)), & \text{if $0 \le s \le \frac{1}{2}$} \\ f(2(1-s)(1-t)), & \text{if $\frac{1}{2} \le s \le 1$} \end{cases}$$
How did they come upon the two functions inside the $f$'s? And how is $I \times I$ being subdivided here? I know the inside functions are functions of the form: $$\theta_t(s) = s \frac{b_1-b_2}{s_1-s_2} + \frac{s_2b_2 - s_2b_1}{s_1-s_2}$$ where $\theta_t : [s_1, s_2] \rightarrow [b_1, b_2]$ is an affine map that maps $s_i$ to $b_i$, but I'm not sure of what the subdivision is for this. Anyone have any ideas?
First, you know that $f^{-1}(s) := f(1-s)$, by definition. Then, the product path $f \star f^{-1}$ is equal to $$ f \star f^{-1} = \begin{cases} f(2s) & s \in [0,\frac12] \\ f^{-1}(2s-1) = f(2(1-s)) & s \in [\frac12,1], \end{cases} $$ again by definition of product path. Note that the endpoints are the same and they equal $p$.
Finally, you need a homotopy (relative to $\dot{I}$). Your additional information is that $i_p(s) = p = f(0)$ over $[0,1]$.
To me, the most simple homotopy to think about, in this case, is
$$ H(s,t) = \begin{cases} f(2s\cdot t) & s \in [0,\frac12] \\ f^{-1}((2s-1)\cdot t) = f(1 - t(2s-1)) & s \in [\frac12,1], \end{cases} $$
which is continuous and satisfies the conditions to be eligible as a homotopy. The homotopy you provided just goes the opposite way: the one I provided satisfies $H(s,0) = i_p$ and $H(s,1) = f \star f^{-1}$.
Note that you can find more than a homotopy, given two functions.