As for the second part, the following works. The goal, inequality (3.4), can be written as $\newcommand{\E}{\mathbb E}$ $$ |\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C(p)\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta} $$ where $X = \sum_{i\in R_t^1}x_i$, etc. The idea is to use the identity $$ ABC - abc = (A-a)BC + a(B-b)C + ab(C - c) $$ with $A=\phi_1(X),B=\phi_2(Y),C=\phi_3(Z)$, and $a = \phi_1(\E X),b=\phi_2(\E Y),c=\phi_3(\E Z)$, and use the triangle inequality, estimating the three addends on the right-hand side using the assumptions on the $\phi_\alpha$. For that, it is helpful to notice, e.g., $$ |\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p, $$ which follows immediately from the first line of Bourgain's proof. Ultimately then, the rest of the argument is using Khinchine's inequality and the conditions on the $\phi_\alpha$ to match the inequality (3.4).
For example, \begin{align*} \E(1+|X-\E X|)^p \le (1+C(p)|x|)^p < c, \end{align*} and so on.