Let $\Psi(n)$ be the number of possibilities to write $n$ as a sum of two coprime composites.
First of all, you ask for a formula for calculating $\Psi$ similar to the formula
$$ \phi({p_1}^{e_1} \cdots {p_k}^{e_k}) = \prod_{i=1}^k \phi({p_i}^{e_i}) = \prod_{i=1}^k p_i^{e_i-1}(p_i-1).$$
Note that this expansion only exists because $\phi$ is multiplicative, i.e., for any two coprime integers $n$ and $m$ it holds that $\phi(nm) = \phi(n) \phi(m)$. However, $\Psi$ is not multiplicative (although the numbers are too big and I don't want to calculate an example right now).
Note that numbers $k$ and $n-k$ are coprime if and only if $n$ and $k$ are
coprime, so the number of ways to write $n$ as the sum of two coprimes is
$\frac 12 (\phi(n)-2)$, where $\phi$ is euler's totient function (we have to
subtract two in order to exclude the representations $n = 1 + (n-1) = (n-1) +
1$). You also
wanted the pairs to consist of composites only, so we need to subtract the
number of primes $p \leq n$ which do not divide $n$, and by
inclusion-exclusion, we need to add the number of ways to write $n$ as a sum of
two primes again. Hence we get
$$ \Psi(n) = \frac 12 (\phi(n)-2) - \pi'(n) + G(n),$$
where
- $\pi'(n)$ is the number of primes $< n-1$ which are coprime to $n$
- $G(n)$ is the number of ways to write $n$ as a sum of two different primes. (This is related to the Goldbach conjecture).
We can trivially bound $\pi'$ and $G$. If $\pi$ denotes the prime counting function, we get
$$ \pi'(n) \leq \pi(n) \ll \frac n {\log n}$$
(you can read $\ll n/\log n$ as $\leq 10 \cdot n /\log n$)
In case where $n$ is odd, we trivially have $G(n) \leq 1$, but in general
$G(n) \leq \pi(n)$ is also good enough.
Finally, it is known that
$$ \liminf \phi(n) \frac{\log \log (n)}{n} = e^{-\gamma},$$
which shows that the approximation
$$ \Psi(n) = \frac 12 \phi(n) + O\left( \frac{n}{\log n}\right)$$
is asymptotic.
We can do a bit better, $\pi'$ should be replacable with $\pi$ at the cost of a small error. Indeed, if $\omega(n)$ denotes the number of prime divisors of $n$, we find $\pi'(n) = \pi(n) - \omega(n)$, and it is easily verified that $\omega(n) \ll \log(n)$, which much smaller than $\pi(n)$. Unfortunately I don't know of a good way to bound $G(n)$ for even $n$. For odd $n$ we obtain
$$ \Psi(n) = \frac 12 \phi(n) - \pi(n) + O(\log n), $$
for even $n$ we can't do better than
$$ \Psi(n) = \frac 12 \phi(n) - \pi(n) + G(n) + O(\log n).$$
In both instances we can also use the prime number theorem in the form
$$ \pi(x) = \mathrm{Li}(x) + O(xe^{-c \sqrt{\log(x)}}).$$
One way or another, the main term is $\frac 12 \phi(n)$, which explains what you observed: If $n$ is prime we find
$$\Psi(n) \approx \frac 12 \phi(n) = \frac 12 (n-1)$$
while for primorials $n = \prod_{i \leq k} p_i$ we find
$$ \Psi(n) \approx \frac 12 \phi(n) = \frac 12 \prod_{i \leq k} (p_i - 1),$$ which gets arbitrarily small compared to $n/2$.
