This may not be a complete answer but is too long for a comment and perhaps provides some insight into the relationship between the twin-prime sieve and the sieve of Eratosthenes.
For the sieve of Eratosthenes the number of elements left after removing all primes up to $p_k$ and their multiples from the set $\left\{1,2,3,...p_k\#\right\}$ is given by
$$E_k=p_k\#\ \prod\limits_{i=1}^k \left(1-\frac{1}{p_i}\right)=\prod\limits_{i=1}^k \left(p_i-1\right)\tag{1}$$
which can be written in terms of the more general Euler totient function
$$\phi(n)=n \prod\limits_{p|n}\left(1-\frac{1}{p}\right)\tag{2}$$
as
$$E_k=\phi\left(p_k\#\right).\tag{3}$$
OEIS entry A066264 gives the number of composites < $p_k\#$ with all prime factors > $p_k$ which corresponds to
$$E_k-1-\left(\pi\left(p_k\#\right)-\pi\left(p_k\right)\right).\tag{4}$$
For the twin-prime sieve the number of elements left after removing all multiples of primes up to $p_k$ and their associates from the set $\left\{1,2,3,...p_k\#\right\}$ is given by
$$N_k=p_k\#\ \prod\limits_{i=2}^k \left(1-\frac{2}{p_i}\right)=2 \prod\limits_{i=2}^k \left(p_i-2\right)\tag{5}$$
which can be written in terms of the more general function
$$\phi_2(n)=n \prod\limits_{p>2\,\land\,p|n}\left(1-\frac{2}{p}\right)\tag{6}$$
as
$$N_k=\phi_2\left(p_k\#\right).\tag{7}$$
I believe $\phi_2(n)$ is related to $\phi(n)$ as follows
$$\phi_2(n)=\sum\limits_{d|n} (-1)^{d-1}\ \mu(rad(d))\ \phi\left(\frac{n}{d}\right)\tag{8}$$
where $\mu(n)$ is the Möbius function and $rad(n)$ is the radical of an integer, so $N_k$ can be computed by evaluating formula (8) above for $\phi_2(n)$ at $n=p_k\#$ which is square-free and therefore $rad(d)=d$ for this case. So $N_k$ is related to $E_k$ as follows:
$$N_k=E_k+\sum\limits_{d>1\,\land\,d\,|\,p_k\#} (-1)^{d-1}\ \mu(d)\ \phi\left(\frac{p_k\#}{d}\right)\tag{9}$$
Note that $\mu(rad(n))=(-1)^{\nu(n)}$ where $\nu(n)$ is the number of distinct primes dividing $n$ (see OEIS entry A001221), so formula (8) above can also be evaluated as follows.
$$\phi_2(n)=\sum\limits_{d|n} (-1)^{d-1}\ (-1)^{\nu(d)}\ \phi\left(\frac{n}{d}\right)\tag{10}$$
When evaluated at $n=p_k\#$, formula (10) above can be simplified to
$$N_k=\phi_2(p_k\#)=2 \sum\limits_{d\,\left|\,\frac{p_k\#}{2}\right.} (-1)^{\nu(d)}\ \phi\left(\frac{\frac{p_k\#}{2}}{d}\right)\tag{11}$$
which combines the contributions of odd and even related divisors $d$ and $2 d$ since
$$(-1)^{2 d-1}\ (-1)^{\nu(2 d)}\ \phi\left(\frac{n}{2 d}\right)=(-1)^{d-1}\ (-1)^{\nu(d)}\ \phi\left(\frac{n}{d}\right)\tag{12}$$
when $(n \bmod 4)=2$ which is the case for $n=p_k\#$.
When $k>1$ formula (11) above can be simplified further as follows
$$N_k=\phi_2(p_k\#)=2 \sum\limits_{d\,\left|\,\frac{p_k\#}{6}\right.} (-1)^{\nu(d)}\ \phi\left(\frac{\frac{p_k\#}{6}}{d}\right)\tag{13}$$
The following table provides insight into the relationship between formula (13) above and the operation of the twin-prime sieve where $k=4$, $p_k=7$, $p_k\#=210$, $m=\frac{pk\#}{6}=35$, $d$ is a divisor of $m$, and $c(k,d)$ in the last column represents the number of times the divisor $d$ appears in a pair of associates when the smallest prime dividing the related associate is greater than $p_k$.
$$\begin{array}{ccc}
d & 2 (-1)^{\nu (d)} \phi \left(\frac{m}{d}\right) & (-1)^{\nu (d)} c(k,d) \\
1 & 48 & \text{-} \\
5 & -12 & -12 \\
7 & -8 & -8 \\
35 & 2 & 2 \\
\end{array}$$
The first row in the Table above removes all primes up to $p_k$ and their multiples equivalent to the sieve of Eratosthenes. What remains is to remove the associates of primes $p_3=5$ up to $p_k$ and their multiples which have no prime divisor $\le p_k$ which is accomplished by the remaining rows in the table. The second row in the table removes the associates of $p_3=5$ and it's multiples which have no prime divisor $\le p_k$. The third row in the table removes the associates of $p_4=7$ and it's multiples which have no prime divisor $\le p_k$. The last row in the table compensates for the fact that two associates were removed twice, once by the divisor $p_3=5$, and a second time by the divisor $p_4=7$.
