This is not an answer but providing a perhaps useful formula for $\pi_k(n)$.
First, you'll need to pick a definition of $\pi_k(n)$ to work with, otherwise you'll be chasing up blind alleys.
I recommend this most simple one. Note that $0$-separated twin prime averages are just primes themselves, so that $k=1$ separated primes are twins and so on.. This formula generalizes all of the cases! Define $\pi_k(n)$ to be the number of $2k$-separated prime pair averages that lie in the range $[1, n]$.
We have, recursively:
$$
\pi_k(n) = \left(\sum_{d\ \mid\ \sqrt{n+k}\#}\mu(d)\left(\sum_{0 \leq r \lt d, \ \\ r^2 = k^2 \mod d}\left\lfloor \frac{n -r}{d}\right\rfloor\right)\right) + \pi_k(\sqrt{n + k} +k ) -1
$$
I.e. recurse until you reach $\pi_k(\sqrt{n + k} + k) = \pi_k(k) = 0$.
Ask me any questions about it and I can teach you those parts that you do not know about.
Note, this is a generalization of an inclusion-exclusion formula on Wikipedia for $\pi_0(n) \equiv \pi(n)$.
Once you realize this formula, you may try plugging it into your conjectures and seeing what you get.
Why it's a good formula: it involves modular arithmetic, group theory, arithmetic functions, inclusion-exclusion, square-free integers, etc. You only need a few of these things to derive it. You don't need group theory for instance, though the groups are just there hiding. It also avoids use of the CRT (Chinese Remainder Theorem). It actually involves a fairly simple to solve inequality in one of its steps.
Edit:
Derivation of Formula
Throughout, $p, q$ will always be prime numbers.
Lemma 1. A number $x \in \Bbb{Z}$ is a twin prime average if and only if $x \pm 1 \neq 0 \pmod q$ for all $2 \leq q \leq \sqrt{ x+ 1}$. Proof. Sieve of Eratosthenes.
Lemma 2. $ab \neq 0 \mod q \iff a \neq 0 \mod q$ and $b \neq 0 \mod q$. This $\iff$ occurs only by primality of $q$, so if $q$ were composite, it wouldn't hold.
Corollary 1. Lemma 1 can be expressed $x$ is a twin prime average if and only if $(x +1)(x-1) \neq 0 \mod q$ for all $2\leq q \leq \sqrt{x + 1}$.
But $(x + 1)(x-1) = x^2 - 1$ more succinctly. Here, if we were studying $2k$-separated prime averages for $k \neq 1$, then this would read $x^2 - k^2 \neq 0 \mod q$.
Thus, given an interval $[1, b] \subset \Bbb{N}$ and that you'd like to count the number $x \in [1,b]$ such that $x^2 \neq 1 \mod q$ what you can do is take the set size :
$$
\# [1,b] - \#\bigcup_{2\leq q \leq \sqrt{x + 1}} \{x\in [1,b] : x^2 =1\mod q\}
$$
Ie. the size of the whole set minus the size of the union of all examples that you don't want to count equals the size that you're interested in.
But what is the size of $\bigcup_{2 \leq q \leq \sqrt{x + 1}} \{x\in [1,b]: x^2 = 1 \mod q\}$?
Here mathematicians usually employ what is called the inclusion exclusion principle.
For example:
$$
\#(A \cup B \cup C) = \# A + \# B + \# C - \# (A\cap B) -\#(A \cap C) - \#(B\cap C) + \#(A\cap B \cap C)
$$
for any three sets $A, B, C$.
But, when you take say $X_q =\{ x \in [1,b ] : x^2 = 1 \mod q\}$ and $X_p = \{x \in \[1,b]: x^2 = 1\mod p\}$ and you intersect them, what do you get?
Each $x \in X_p \cap X_q$ is such that $x^2 = 1\pmod q$ and $x^2 = 1 \pmod p$.
Lemma 3. By the Chinese Remainder Theorem we have that $X_{q_1} \cap \dots \cap X_{q_n}$ for prime numbers $q_i$ is equal to $X_{q_1 \cdots q_n}$.
Now, our inclusion-exclusion formula becomes:
$$
\psi(b) = \sum_{d\ \mid\ \sqrt{b + 1}\#} \# X_d(b)
$$
where we note that $\#[1,b]$ got absorbed by this notation into summation case for when $d = 1$ and certainly divides $\sqrt{b + 1}\#$, because $\#X_1(b) =$ the set of all $x \in [1,b]$ such that $x^2 = 1 \mod 1$, but everything zeros out modulo $1$, so that equation is true for all $x \in [1,b]$ and therefore $\#X_1(b) = b$.
Now how can we expand or express $\#X_d(b)$? One way is to note that what you're trying to solve is:
$$
1 \leq x = dz +r \leq b
$$
where $0 \leq r \lt d$ and solves $r^2 = 1 \mod d$. These are all the different $x$'s modulo $d$ that make up the count in $\#X_d(b)$.
Since an $x$ can belong to one-and-only-one equivalence class modulo $d$, we have that summation is appropriate, and that $$\#X_d(b) = \sum_{0\leq r \lt d, \\ r^2 = 1 \mod d} \left\lfloor\frac{b - r}{d} \right\rfloor$$
It's like saying yes, there are some $x \in [1, d)$ that solve it but also some in $[d, 2d)$ and son on... So you've got to count "how many copies of $[0,d)$" can we squeeze into $b - r$ and that's $\lfloor\frac{b - r}{d} \rfloor$.
To re-iterate, since $x = dz + r$ solves $1 \leq x \lt b$ and $x^2 = 1 \mod d$, how many more times does $x$ (modulo $d$) occur in the interval $[1,b]$ and that's where there the division-floor expression comes from.
General Formula:
$$
\psi_k(b) = \sum_{d\ \mid\ \sqrt{b + k}\#}\mu(d)\left(\sum_{0 \leq r \lt d, \ \\ r^2 = k^2 \mod d} \left\lfloor\frac{b - r}{d} \right\rfloor\right)
$$
But as the formula includes a prime $q \mid \sqrt{b + 1}\#$ , if that is a twin prime average, it gets "cancelled" from the count. Thus we must add in, recursively:
$$
\pi_k(b) = -1 +\pi_k(\sqrt{b + k } + k) +\sum_{d \ \mid \ \sqrt{b + k}\#} \mu(d) \left(\sum_{0\leq r \lt d, \ r^2 = k^2 \mod d } \left\lfloor\frac{b - r}{d}\right\rfloor\right)
$$
By Sieve of Eratosthenes, $\psi_k(b)$ will count what we want in the interval
$\sqrt{b + k} + k + 1$ to $b$, so what we do is recursively count the rest in the first term of $\pi_k(b)$'s definition.
Where the $-1$ comes from it's a correction from something, but I can't figure out where. It shows up when you're code verifying the formula - everything is off by 1, so I just subtract 1. Also the Wikipedia article has a 1 there. So for it to "align with" that, I must include it. It either is there or it isn't, but to be on the safe side I'll include it.
Anyways, that's how you derive the formula.
Now prove that it is not eventually constant. That's the hard part.
