You could use a heuristic just based on the number of odd primes: The chance that a large odd n is prime is roughly 2 / ln n. You can estimate the chances that k consecutive large integers are all primes. You can then calculate for example “if I check all even integers from 10^(j-1)to 10^j, and given k, what are the chances that there is any n in that range so that n - p is non-prime for each of the first k primes”.
You get a bit more precision if you take into account small divisors of n. If n is a multiple of 3, for example, then n-3 is composite instead of a 2/ln n chance of being prime. But for all other p, n-p is not divisible by 3, making it a lot more likely that n-p is prime (3 / ln n instead of 2 / ln n).
On the other hand, if n is not a multiple of 3, then n-p for p≠3 is more likely to be divisible by 3 then a random large integer. So n which are not divisible by 3, 5, 7 are more likely to produce a longer sequence of primes with n-p composite. That will make your first heuristic underestimate the maximum. If you look at your numbers, none above 30 are divisible by 3, and only one after 220 is divisible by 5.
If n is not divisible by some (small) prime q, and we take a random prime p > 2, p ≠ q, then there are q-1 possible values n modulo q, and q-1 possible values p modulo q. n-p is divisible by q in one in q-1 cases instead of one in q cases if n-p was a random integer.
The probability that n-p is prime for large even n, odd p, would usually be 2 / ln n. (1 / ln n for random integers, twice as high because n-p is odd). If n is divisible by q, then the chance that n-p is NOT divisible by q is (q-2) / (q-1) instead of (q-1) / q for random numbers, so the chance that n-p is prime is multiplied by (q^2 - 2q) / (q-1)^2 = 1 - 1 / (q-1)^2. This factor is 3/4 for q = 3, 15/16 for q = 5, 35/36 for q = 7, 120/121 for q = 11. The product of these factors for many q is about 0.66. So the chance that n-p is prime is not 2 / ln n but about 1.32 / ln n. On the other hand, there are fewer n which are not divisible by 3, 5, 7, 11 etc.
If the chance of n-p being prime is 1.32 / ln n, the chance of n-p being composite = 1 - 1.32 / ln n, then the chance that n-p is composite for k different p is about (1 - 1.32 / ln n)^k ≈ exp(-1.32k / ln n), the chance that one p is the matching prime for n is ≈ 1 - exp (1.32k / ln n). If we check M values for n, the chance that we always find a matching prime among the first k primes is (1 - exp (1.32k / ln n))^M. For large k, this is about exp (-M * exp (1.32k / ln n)). (Haven't tested these formulas, so there might be mistakes).
If you decide on say M = n / 2 ln n, which is roughly the number of primes from n/2 to n, and decide you want a probability of say 0.5, then you can calculate k easily. For example: ln 0.5 = -M exp (1.32k / ln n), ln 2 * n / 2 ln n = exp (1.32k / ln n), 1.32 k / ln n = ln (ln 2 * n / 2 ln 2n), k = ln n * ln (n ln 2 / 2 ln 2n) / 1.32. Very roughly.
And that's the number of primes. The k'th prime is very roughly k ln k. So the prime required would be about c * (ln n)^2 * ln ln n and c = 1.2 gives quite reasonable results (the $p_n$ are not a very smooth function).
This is not a complete and rigorous answer, but I think it is a good start to an explanation.
Up to $x$, you can squeeze in more numbers with $3$ as a factor than you can numbers with $5$ as a factor. Just from $3$ and $5$, you have a difference of something like $\frac{1}{3}-\frac{1}{5}=0.1\overline{3}$.
Trying to measure this, I believe what you have is:
$$f(x)=\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{7}\right\rfloor+\left\lfloor\frac{x}{11}\right\rfloor+\cdots$$
$$g(x)=\left\lfloor\frac{x}{5}\right\rfloor+\left\lfloor\frac{x}{13}\right\rfloor+\left\lfloor\frac{x}{17}\right\rfloor+\cdots$$
So $\frac{f(x)-g(x)}{x}$ is something like:
$$\frac13-\frac15+\frac17+\frac1{11}-\frac1{13}-\frac1{17}+\frac1{19}+\cdots$$
This is converging to something that is suspiciously close your result ($0.3349\ldots$) when only examining odd numbers less than $x$.
Note that your result for looking among even numbers ($1.3349\ldots$) is this number plus $1$. And then the original of your three results ($0.8349\ldots$) is the average of those two numbers. Of course averaging the first two numbers would make sense, but why is the even count 1 more than the odd count?
Are factors of $2$ being properly omitted from your counts for $f(x)$ and $g(x)$? I notice that $\frac12+\frac14+\frac18+\cdots=1$, and if factors of $2$ (with repetition) were included in the count for $f(x)$, it would account for that additional $1$. But I am just speculating.
Best Answer
I expect that $$\sum_{p \leqslant x} f(p) = x\log \log x - x\log \log \log x + O(x)\,, \tag{$\ast$}$$ but I don't see how that could be proved without knowing much stronger bounds on prime gaps than we currently do. Since $\log \log \log x$ grows very very slowly, this would not easily be distinguished from $x\log \log x - x$ empirically.
It is not difficult to show that $$\sum_{p \leqslant x} f(p) \leqslant x\log \log x - x\log \log \log x + C\frac{x}{\log \log x} \tag{1}$$ for a suitable constant $C$ using the known bounds for prime gaps. Proving lower bounds is harder.
To estimate the sum, let's "switch the order of summation". Instead of counting the number of primes having a multiple in each composite run (the composite numbers between two successive primes), for each prime count the number of consecutive runs starting at or below $x$ in which the prime has a multiple.
Things are easier to write down if we consider only the multiples $\leqslant x$. This doesn't make a difference for $(1)$, since by a result of Hoheisel subsequently improved by various people, the length of the last composite run to be considered is at most $x^{\theta}$ for some $\theta < 1$. By the trivial bound $\omega(n) \ll \log n$, ignoring the numbers $> x$ in that run introduces an $O(x^{\theta}\log x)$ error, comfortably smaller than the $O\bigl(\frac{x}{\log \log x}\bigr)$ term in $(1)$.
Then for each prime $p \leqslant x$, the number of composite runs in which it has a multiple that we count is bounded above on the one hand by $\pi(x)-1$ (since there are at most that many nonempty runs we consider), and on the other hand by $\bigl\lfloor \frac{x}{p}\bigr\rfloor - 1$ since $p$ has just that many multiples $\leqslant x$ excepting $p$ itself. Taking the first bound for small primes and the second one for larger ones, we obtain (for not too small $x$) \begin{align} \sum_{p \leqslant x} f(p) &\leqslant \sum_{p \leqslant \log x} \bigl(\pi(x)-1\bigr) + \sum_{\log x < p \leqslant x} \biggl(\biggl\lfloor \frac{x}{p}\biggr\rfloor - 1\biggr) + O\bigl(x^{\theta}\log x\bigr) \\ &\leqslant \pi(x)\pi(\log x) + x \sum_{\log x < p \leqslant x} \frac{1}{p} + O\bigl(x^{\theta}\log x\bigr) \\ &= x\biggl(\log \log x - \log \log \log x + O\biggl(\frac{1}{\log \log x}\biggr)\biggr) + \pi(x)\pi(\log x) + O\bigl(x^{\theta}\log x\bigr) \\ &= x\log \log x - \log \log \log x + O\biggl(\frac{x}{\log \log x}\biggr) \end{align} by Mertens's second theorem and the Chebyshev bounds. (And we can by these means find an explicit $C$ ifwe wish to do so.)
In order to discuss lower bounds for the sum, let $G(x)$ denote the largest prime gap for which the smaller prime doesn't exceed $x$. Then it is clear that for primes $p > G(x)$ the number of composite runs in which $p$ has a multiple is precisely the number of composite multiples of $p$ not exceeding $x$ (plus maybe one), since such a prime cannot have more than one multiple in a single run. Hence we have $$\sum_{p \leqslant x} f(p) \geqslant \sum_{G(x) < p \leqslant x} \biggl(\biggl\lfloor \frac{x}{p}\biggr\rfloor - 1\biggr) = x\log \log x - x \log \log G(x) + O\biggl(\frac{x}{\log G(x)}\biggr)\,.$$ If, as is widely believed, we have $G(x) \in O\bigl((\log x)^k\bigr)$ for some exponent $k$ (the case $k = 2$ is Cramér's conjecture), then $\log \log G(x) = \log \log \log x + O(1)$, and $(\ast)$ follows. If on the other hand $G(x)$ can be as large as $x^{\varepsilon}$ for some $\varepsilon > 0$, then the arguments above aren't even sufficient to establish the principal term $x\log \log x$.