Very rough heuristic of the $k$th such $n$ (prime signature twin) as a function of $k$:
1st assumption: The dominant effect is given by primes of the signature $(1,1)$, i.e. by semiprimes.
2nd assumption: The number of semiprimes below $n$ is given by $\pi_2(n) \sim \frac{n}{\ln n} \ln \ln n$, cf. http://en.wikipedia.org/wiki/Almost_primehttps://en.wikipedia.org/wiki/Almost_prime
Then the probability density of the semiprimes is approximately given by $f_2(n) \sim \frac{\ln \ln n}{\ln n}$.
3rd assumption: The semiprimes are independently distributed.
Then the number of semiprime twins below $N$ is given by $\int^{N} f_2(x)^2 dx$.
Thus the number of prime signature twins is rouglyroughly given by $(\frac{\ln \ln n}{\ln n})^2$ (a better value or an asymptotic formula can be obtained by evaluating the integral.)
Thus the $k$th prime signature twin is roughly given byh $n = k \cdot (\frac{\ln k}{\ln \ln k})^2$. For $k = 200$ (as in the figure cited by the OP) this gives approximately a slope of 8,8, the same order of magnitude as in the figure.
Of course this very rough calculation can be improved in various ways.
