Let $X_1, X_2, \dots$ be i.i.d.; $\mathbb{E} [X_i] = 0$; $\mathbb{E} [e^{t X_i}] \le e^\frac{t^2}{2}$ for $t \in \mathbb{R}$. It can be shown, by Markov's inequality, that $\mathbb{P} \left[ \sum_{i = 1}^n X_i \ge s \right] \le e^{-\frac{s^2}{2n}}$, where $t = \frac{s}{n}$.
From there, the proof argues, with $s := qn, q >0$, that $\mathbb{P} \left[ \frac{1}{n} \left| \sum_{i = 1} X_i \right| \ge q \right] = \mathbb{P} \left[ \frac{1}{n} \sum_{i = 1} X_i \ge q \right] + \mathbb{P} \left[ -\frac{1}{n} \sum_{i = 1} X_i \ge q \right] \le 2 e^{-\frac{n q^2}{2}}$, to then sandwich the former probability with Borel–Cantelli.
I understand that $\{ \frac{1}{n} \left| \sum_{i = 1} X_i \right| \ge q \} = \{ \frac{1}{n} \sum_{i = 1} X_i \ge q \} \cup \{ -\frac{1}{n} \sum_{i = 1} X_i \ge q \}$ is a disjoint union and that $\mathbb{P} \left[ \frac{1}{n} \sum_{i = 1} X_i \ge q \right] \le e^{-\frac{n q^2}{2}}$. How can I argue, though, that also $\mathbb{P} \left[ -\frac{1}{n} \sum_{i = 1} X_i \ge q \right] \le e^{-\frac{n q^2}{2}}$ (which the proof seems to do)?
Very simple! You already know that $P[\frac 1 n \sum_{i=1}^{n} X_i \geq q] \leq e^{-nq^{2}} /2$ whenever $\{X_i\}$ is i.i.d with mean 0 and $Ee^{tX_i} \leq e^{t^{2}/2}$ for all real $t$. Just change $\{X_i\}$ to $\{-X_i\}$ in this and verify that the hypothesis is still satisfied.