I was reading Professor Nathanson's graduate texts in mathematics 164: "Additive Number Theory The Classical Bases" (http://www.alefenu.com/libri/nathansonbases.pdf), and I was wondering if I may ask two questions about Weights in chapter 10.2 (p. 272).
Choosing optimal weights in sieve theory is a very difficult problem that is often done by trial and error. In Nathanson's book it seems as if he was attempting to produce the simplest and shortest version of Chen's theorem. Whereas, in another book, such as Halberstam and Richert's "Sieve methods", weighting procedures are introduced more gradually and consequently allow the reader to get a better feel for what is optimal.
To answer your first question, the choices of $z=N^{1/8}$ and $y=N^{1/3}$ can be modified slightly. However, the exponents here affect the sieving argument and bad choices will not give $r(N)>0$. Nathanson's weights are certainly not optimal, and choosing $z=N^{1/7.9}$ say, may give a better lower bound for $r(N)$. If you look at a paper such as [1], then you'll see how complicated the choice of exponents can get.
I will attempt to roughly explain how Nathanson's choice of sieve weights works. I'll start with his final expression (Theorem 10.2) which is obtained from $w(n)$:
$$ r(N)>S(A,P,z)-\frac{1}{2}\sum_{z\leq q<y}S(A_q,P,z)-\frac{1}{2}S(B,P,y) + O(N^{7/8})\qquad (*).$$
Firstly, $S(A,P,z)$ sieves out all elements of A with prime factors less than $z$. Thus, if $z=N^{1/u}$ for some integer $u$, $S(A,P,z)>0$ implies that every large even $N$ can be written as the sum of a prime and a number with at most $u-1$ prime factors.
Secondly, if $y=N^{1/v}$ for some integer $v$, the sum $$-\frac{1}{2}\sum_{z\leq q<y}S(A_q,P,z)$$ essentially sieves out all of the elements left in $A$ except those with at most $v$ prime factors. This is explained more precisely by Nathanson in Section 10.2. Note that this sum could be more generally replaced by $$-\frac{1}{M}\sum_{z\leq q<y}S(A_q,P,z),$$ which would instead leave you with the elements of $A$ with at most $v+(M-2)$ prime factors.
At this stage, the choice of $u=8$ and $v=3$ allows us to get close to Chen's theorem in that the first two terms of $(*)$ only leave us with elements of $A$ with at most 3 prime factors. Note that it is important to split this argument into two parts as just setting $u=4$, one does not get $S(A,P,z)>0$. The key thing here is that estimating the sum over $S(A_q,P,z)$ allows us to utilise a sieve upper bound (rather than just a lower bound) which gives a better result.
The final term
$$-\frac{1}{2}S(B,P,y)$$
is then the key novelty in the proof of Chen's theorem, as it allow us to go from 3 prime factors to 2 prime factors by utilising a bilinear form estimate.
I hope this helps. As Stanley Xiao suggested in the comments, it could be useful exercise to try to repeat the argument in Nathanson's book with different choices of $z$ and $y$ and see what gives the best result.
[1] Y. Cai, On Chen's theorem (II), J. Number Theory, 2008.
