Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$: $$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$ Then $s\circ f_{a,b}$ will have all but finitely many blocks of 1s of sizes $\lfloor2^{2^{k}}/a\rfloor$ and $\lceil 2^{2^k}/a\rceil$.

We claim that we cannot keep having $k>j$ and $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ (and therefore all $s\circ f_{a,b}$ are distinct).

Indeed, when this happens then $2^{2^k-2^j}\le a_1/{a_2}+a_1 2^{-2^j}$ is bounded. But as $k>j\to\infty$, $2^{2^k-2^j}$ is unbounded.

(Note that $2^k/a_1=2^{k-1}/a_2$ with $a_1=4$ and $a_2=2$, so a single-exponential $2^k$ in place of a double-exponential $2^{2^k}$ is not enough for this construction.)

Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$: $$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$ Observe that all but finitely many blocks of 1s of $s\circ f_{a,b}$ have size of the form $\lfloor2^{2^{k}}/a\rfloor$ or $\lceil 2^{2^k}/a\rceil$.

We claim that we cannot keep having $k>j$ and $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ (and therefore all $s\circ f_{a,b}$ are distinct).

Indeed, when this happens then $2^{2^k-2^j}\le a_1/{a_2}+a_1 2^{-2^j}$ is bounded. But as $k>j\to\infty$, $2^{2^k-2^j}$ is unbounded.

(Note that $2^k/a_1=2^{k-1}/a_2$ with $a_1=4$ and $a_2=2$, so a single-exponential $2^k$ in place of a double-exponential $2^{2^k}$ is not enough for this construction.)

Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$: $$01\, 0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$ Then $s\circ f_{a,b}$ will have all (but one) blocks of 1s of sizes $\lfloor2^{2^{k}}/a\rfloor$ and $\lceil 2^{2^k}/a\rceil$.

If $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ then $2^{2^k-2^j}\le a_1/{a_2}+2^{-2^j}$. But if $k>j\to\infty$ then $2^{2^k-2^j}$ is unbounded.

(Note that $2^k/4=2^{k-1}/2$, so $2^k$ in place of $2^{2^k}$ is not enough for this construction.)

In conclusion, all $s\circ f_{a,b}$ are distinct.

Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$: $$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$ Then $s\circ f_{a,b}$ will have all but finitely many blocks of 1s of sizes $\lfloor2^{2^{k}}/a\rfloor$ and $\lceil 2^{2^k}/a\rceil$.

We claim that we cannot keep having $k>j$ and $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ (and therefore all $s\circ f_{a,b}$ are distinct).

Indeed, when this happens then $2^{2^k-2^j}\le a_1/{a_2}+a_1 2^{-2^j}$ is bounded. But as $k>j\to\infty$, $2^{2^k-2^j}$ is unbounded.

(Note that $2^k/a_1=2^{k-1}/a_2$ with $a_1=4$ and $a_2=2$, so a single-exponential $2^k$ in place of a double-exponential $2^{2^k}$ is not enough for this construction.)