But if Hooley R can only be applied to one n-variable, the result is HN_1 * N_2*sqrt(N_3), and this is x^(1/2+5/16+5/32) in the border case, not beating x^(15/16).

Instead, he seems to take it as small as possible (13.12) to beat the losses incurred from $H$ introduction and/or $N_1$ exceeding of $x^{3/8}$ and $d_1$ exceeding of $x^{1/2}$ (namely $8\omega+2\omega$, which is $40\omega$ as you win by $\sqrt r$ and apply Cauchy, these $N_1$ and $d_1$ are linked to $H$ selection so you can see it either way).

Unless I am wrong, this is an inefficiency. I think $r$ can be something like $x^{1/12-6\omega}$ or $x^{1/10-8\omega}$ for 14.8 and 14.5, also 14.4 to work.

I think a philosophical reason partially as to why it works is that his Weyl shift was by $hkr$, notably a multiple of $r$ so that the Ramanujan sum pops out of the factorization $e_d()$ down to $e_r()e_s()$. But I agree that is mysterious.