The following is a slight oversimplification, but I think it answers the heart of your question: have a look at Definition 9.2 in Green-Tao:

$$\Lambda_R(n) := \sum_{\substack{d\mid n\\d \le R}} \mu(d) \log(R/d).$$

Also note in Definition 9.3 that $R$ is taken as a small positive power $N^{c_k}$, where $[1,N]$ is (more or less) the interval in which one is sieving for arithmetic progressions.

The function $\Lambda^2_R(n)$, gets stretched and chopped in a few ways by Definition 9.3, but it's essentially the concrete measure you're looking for. Notice that when the smallest prime factor of $n$ is at least $R$, then $\Lambda_R(n) = \log R$: these turn out to be (the bulk of) the numbers for which the measure is spikiest. By contrast, if $n$ has some prime factors that are smaller than $R$, then we'll get some cancellation from the sign changes of $\mu$.

While it's theoretically conceivable that some highly composite values of $n$ yield sums which conspire to larger values of $\Lambda_R$, this is just not something that happens often enough to matter. Indeed, the average magnitude of $\Lambda_R$ over a suitably long interval is just $O(1)$, and the average value of $\Lambda_R^2$ is $O(\log R)$ (the point of Section 10 is to prove a much stronger form of this), so that the contribution from just the primes in $[R,N]$ has positive density with respect to the measure.

Although Chen's theorem is usually stated in the simple language of semiprimes, the proof actually yields something quite a bit stronger (see for instance p.483 of Opera de Cribro): there are infinitely many primes $p$ such that $p-2$ has at $\le 2$ prime divisors, each of which is at least $p^{1/8}$.

To make sense of Green's comparison to Chen primes, I think it's reasonable in light of the previous paragraph to approximate "semiprimes" by "numbers $n$ whose prime factors are all larger than $n^{c_k}$", a condition that is in some ways weaker (it allows more than 2 prime factors) and in some ways stronger (they are restricted in size). Most significantly, this addresses your concern about the cardinality of semiprimes. While you are of course correct that there are $N \log \log N / \log N$ semiprimes up to $N$, this does not apply to the second set which is only of size $O(N/\log N)$ (see Buchstab's theorem) and thus comparable to the primes themselves.


The sequence given (I did not cross-check with the presentation) seems to consist of numbers which are product of two distinct primes, both greater than $3$.