Python 3, 78 77 75 70 68 62 bytes
f=lambda n,k=3,m=1,j=0:k<n and-m%k*j*2/k+f(n,k+2,m*k**4,m%k/k)
Thanks to @xnor for golfing off 2 4 bytes and paving the way for 4 more!
Try it online!
Background
Recall that Wilson's theorem states that for all integers k > 1,
where a ≡ b (mod d) means that a - b is evenly divisible by d, i.e., a and b have the same residue when divided by d.
In Wilson Theorems for Double-, Hyper-, Sub- and Super-factorials, the authors prove generalizations for double factorials, on which this answer builds. The double factorial of an integer k ≥ 0 is defined by
Theorem 4 of the aforementioned paper states the following.
Elevating both sides of the congruences to the fourth power, we deduce that
for all odd primes p. Since 1!! = 1, the equivalence holds also for p = 2.
Now, doing the same to Wilson's theorem reveals that
Since
it follows that
whenever p is prime.
Now, let k be an odd, positive, composite integer. By definition, there exist integers a, b > 1 such that k = ab.
Since k is odd, so are a and b. Thus, both occur in the sequence 1, 3, …, k - 2 and
where | indicates divisibility.
Summing up, for all odd integers k > 1
where p(k) = 1 if k is prime and p(k) = 0 if k is composite.
How it works
When the function f is called with a single argument, k, m, and j are initialized as 3, 1, and 0.
Note that ((k - 2)!!)4 = 1!!4 = 1 = m. In fact, the equality m = ((k - 2)!!)4 will hold at all times. j is a float and will always be equal to ((k - 4)!!)4 % (k - 2) / (k - 2).
While \$k < n\$, the right argument of and will get evaluated. Since \$j = \frac{\mod(\left(\left(k - 4\right)!!\right)^4 , k - 2)}{k - 2}\$, as proven in the first paragraph, \$j = \frac{1}{k - 2}\$ if \$k - 2\$ is prime and \$j = 0\$ if not. Likewise, since \$\mod(m,k) = \left(\left(k - 2\right)!!\right)^4\$ equals \$1\$ if \$k\$ is prime and \$0\$ if not, \$\mod(-m,k) = k - 1\$ if \$k\$ is prime and \$\mod(-m , k) = 0\$ if not. Therefore, -m%k*j*2/k evaluates to \$\frac{2\left(k - 1\right)}{k\left(k - 2\right)} = \frac{\left(k - 2\right) + k}{k(k - 2)} = \frac{1}{k} + \frac{1}{k - 2}\$ if the pair \$\left(k - 2, k\right)\$ consists of twin primes and to \$0\$ if not.
After computing the above, we add the result to the return value of the recursive call f(n,k+2,m*k**4,m%k/k). \$k\$ gets incremented by \$2\$ so it only takes odd values‡†, we multiply \$m\$ by \$k^4\$ since \$mk^4 = \left(\left(k - 2\right)!!\right)^4k^4 = \left(k!!\right)^4\$, and pass the current value of \$\frac{\mod(m,k)}{k}\$ – which equals \$\frac{1}{k}\$ if the "old" \$k\$ is a prime and \$0\$ if not – as parameter \$j\$ to the function call.
Finally, once \$k\$ is equal to or greater than \$n\$, \$f\$ will return False and the recursion stops. The return value of \$f(n)\$ will be the sum of all \$\frac{1}{k} + \frac{1}{k - 2}\$ such \$\left(k - 2, k\right)\$ is a twin prime pair and \$k < n\$, as desired.
‡ The results from the Background paragraph hold only for odd integers. Since even integers cannot be twin primes, we can skip them safely.
