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As she writes, first apply Cauchy-Schwarz, and only then apply the large sieve (twice). The relevant instance of Cauchy-Schwarz is $$|x_1 x_2| \le \frac{|x_1|^2+|x_2|^2}{2},$$ which, by replacing $x_1$ and $x_2$ by $x_1\sqrt{C}$ and $x_2/\sqrt{C}$ ($C>0$) becomes $$|x_1 x_2| \le \frac{C |x_1|^2 + C^{-1} |x_2|^2}{2}.$$ We apply it with $x_1=\sum_{a \in \mathcal{A}} \chi(a)$ and $x_2 = \sum_{b \in \mathcal{B}}\chi(b)$ and with $C$ to be determined later (but independent of $\chi$). We obtain that the relevant sum is $$\le \frac{1}{2}\left( C\sum_{q \le Q} \frac{q}{\phi(q)}\sum_{\substack{\chi \bmod q\\ \text{primitive}}}\left|\sum_{a \in \mathcal{A}} \chi(a)\right|^2 + C^{-1}\sum_{q \le Q} \frac{q}{\phi(q)}\sum_{\substack{\chi \bmod q\\ \text{primitive}}} \left|\sum_{b \in \mathcal{B}} \chi(b)\right|^2 \right),$$ which, by two applications of the large sieve, is $$\le \frac{1}{2}(Q^2 + N)| \left( CA+ C^{-1} B\right).$$ Now take $C=\sqrt{B/A}$.