I think can be shown as follows as well. Write $$c_r = \overline{\left(\sum_{n\le N} e(n^2t_r)\right)}$$ (where we write $e(t) = e^{2\pi i t}$ for convenience. Then, we obtain that $$\delta^2 NR < \sum_r c_r\sum_n a_n e(n^2t_r) = \sum_{n}a_n\sum_r c_re(n^2t_r)\le\left(\sum_n \bigg|\sum_r c_re(nt_r)\bigg|^2\right)^{1/2} $$ by Cauchy-Schwarz. Then, since we have that $|c_r| > \delta N^{1/2}$, expanding out the square we obtain that $$\delta^4N^2R^2\le \sum_{r, r'} c_r\overline{c_{r'}}\sum_{n} e(n^2(t_r - t_{r'})) \le \delta^2 N\sum_{r, r'}\bigg|\sum_{n} e(n^2(t_r - t_{r'})\bigg| $$ so (4.12) follows by rearranging. I'm not sure if this is the linearisation in the sense Bourgain means.

I think it can be shown as follows. Write $$c_r = \overline{\left(\sum_{n\le N} e(n^2t_r)\right)}$$ (where we write $e(t) = e^{2\pi i t}$ for convenience. Then, we obtain that $$\delta^2 NR < \sum_r c_r\sum_n a_n e(n^2t_r) = \sum_{n}a_n\sum_r c_re(n^2t_r)\le\left(\sum_n \bigg|\sum_r c_re(nt_r)\bigg|^2\right)^{1/2} $$ by Cauchy-Schwarz. Then, since we have that $|c_r| > \delta N^{1/2}$, expanding out the square we obtain that $$\delta^4N^2R^2\le \sum_{r, r'} c_r\overline{c_{r'}}\sum_{n} e(n^2(t_r - t_{r'})) \le \delta^2 N\sum_{r, r'}\bigg|\sum_{n} e(n^2(t_r - t_{r'})\bigg| $$ so (4.12) follows by rearranging. I'm not sure if this is the linearisation in the sense Bourgain means.