Proposition. every map $\alpha: [s_0,s_1] \to \mathbb{R}^n$ is homotopic rel $\{s_0,s_1\}$ to the linear map
$$\beta=\frac{(s_1-s_0)\alpha(s_0)+(s-s_0)\alpha(s_1)}{s_1-s_0}:[s_0,s_1] \to \mathbb{R}^n$$
The proof says
Proof. Define a rel $\{0,1\}$ homotopy $\gamma:\alpha \simeq\beta$ by
$$\gamma(s,t)=(1-t)\alpha(s)+\beta(s):I \times I \to \mathbb{R}^n$$
Question marks all over this one.
Firstly, in this context of the problem, I understand homotopic rel $\{0,1\}$ to be some homotopy $\gamma$ that satisfies
$$\gamma(0,t)=\alpha(0)=\beta(0) \text{ and } \gamma(1,t)=\alpha(1)=\beta(1)$$
for any $t \in I$. Am I right so far?
Then if I check with the $\gamma$ given in the proof, I am not convinced it's a homotopy rel $\{0,1\}$. I mean, let us try $s=0$.
$\gamma(0,t)=(1-t)\alpha(0)+t\beta(0)$. I do not know what $\alpha(0)$ is since $\alpha(s)$ is not specified, so leave that as it is. $\beta(0)$ is computable so substitute $s=0$ and $s_0=0,s_1=1$ in the above given $\beta(s)$ and I get $\beta(0)=\alpha(0)+\alpha(1)$.
Therefore, $\gamma(0,t)=(1-t)\alpha(0)+t(\alpha(0)+\alpha(1))=\alpha(0)-t\alpha(0)+t\alpha(0)+t\alpha(1)=\alpha(0)+t\alpha(1)$
$$\gamma(0,t)=\alpha(0)+t\alpha(1)$$
Well, is $\alpha(0)+t\alpha(1)=\alpha(0)=\beta(0)$ for all $t \in I$? It doesn't look like it, and even if it is, why is it equal??
This is extremely obscure to me, I simply don't understand it, how is $\gamma$ a homotopy rel $\{0,1\}$ of $\alpha, \beta$?
This should be a comment, not an answer, but I'm new here.
There are two typos that seem to be causing you problems. First, your formula for $\beta$ should read $$\beta(s) = \frac{(s_1-s)\alpha(s_0)+(s-s_0)\alpha(s_1)}{s_1-s_0} : [s_0, s_1] \to \mathbb{R}^n.$$ I've changed an $s_0$ to an $s$, which should fix your calculation for $\beta(s_0)$.
Second, and more fundamentally - this may be a typo in your book - your homotopy should be rel $\{s_0, s_1\}$, not rel $\{0, 1\}$.
Then you should be able to check $\beta(s_0) = \alpha(s_0)$, so that $$\gamma(s_0, t) = (1-t)\alpha(s_0) + t\beta(s_0) = \alpha(s_0) = \beta(s_0)$$ as desired, and likewise for $s_1$. But that first typo seems to be screwing you up.