The first paper referenced by OP is basically talking about the Buchstab's identity in a general context compared to Buchstab's application of it in Goldbach's problem, but it is still more specific than the version presented in modern texts such as Halberstam & Richert's Sieve Methods and Friedlander & Iwaniec's Opera de Cribro.

They are regarded as 2 different approaches to number-theoretic problems

By the way, I have written a Chinese article explaining the application of circle method to Goldbach's problem. Hope you will find that one helpful as well.

@joro That infinite product expansion follows from Hadamard's factorization theorem, which allows you to construct entire functions with prescribed zeros.

As a result, the strategy is to give asymptotic estimates for integral over major arcs (collection of these small neighborhoods of rationals) and to give upper bound for integral over minor arcs (i.e. the integral over region in $[0,1]$ that are relatively distant from rationals)

The motivation is that the function $f(\alpha)$ has good properties when $\alpha$ is close to some rational number, and it is also true that the integral over some small neighborhoods of rationals turn out to be the constituents of the main term in the asymptotic formula.

It would be better if you clarify the definition of $D(N)$

I'm not sure whether a literature is available for that, but the identity $p^2-1=p^2(1-p^{-2})$ allows one to get something like $$\sum_{d>z}{\mu^2(d)\over d^2}\prod_{p|d}(1-p^{-2})^{-1}<\zeta(2)\sum_{d>z}{\mu^2(d)\over d^2}$$

$H=\zeta(2)-\sum_{d>\sqrt D}{\mu^2(d)\over\prod_{p|d}(p^2-1)}$, and the latter sum might allow you to replace $3\cdot2^{-1/3}$ with a better constant.

@HAHelfgott The result appears in exercise 6 of section 3.2 of Montgomery & Vaughan's Multiplicative Number Theory I: Classical Theory.

Euler product expansion gives $\sum_{d\ge1}{\mu^2(d)\over\prod_{p|d}(p^2-1)}=\prod_p\left(1+{1\over p^2-1}\right)=\zeta(2)$, so there is no need to perform the ordinary lower-bounding trick here.

As this result is unpublished, I still put it here. Selberg's sieve allows one to conclude the number of squarefree integers in $(x,x+u]$ is $\le u/\zeta(2)+O(u^{2/3})$

@StevenClark Elementary arguments show that $\sum_{n\le x}|\mu(n)|=x/\zeta(2)+O(\sqrt x)$, so there is no need for RH.

I just read the Polymath paper that improves the gap to 246, and I found out in page 11 that the authors are also using the weights equivalent to $v_n$ in my question.