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Added 30-08-2024: it turns out that a forty-years old paper by Murray Bell contains all the ingredients to answer the original question a sequence $\langle B_n:2\le n<\omega\rangle$ of subalgebras of $\mathcal{P}(\omega)/\mathit{fin}$ such that $B_n$ is $\sigma$-$n$-linked but not $\sigma$-$(n+1)$-linked (for all $n$). This can be converted into a sequence as desired in the question. Details in the paper on ArXiV.org.

16-03-2023: I put a more leisurely explanation on ArXiV.org

As above: let $x\in C(f)\cap A$ and assume $f(x)\in V$. Then $s(x)=\langle f(x),0\rangle$ or $s(x)=\langle f(x),1\rangle$. As above we split $C(f)\cap A$ into two pieces: $C_0=\{x:s(x)=\langle f(x),0\rangle\}$ and
 $C_1=\{x:s(x)=\langle f(x),1\rangle\}$.

If $x\in C_0$ then we take a rational interval $(p_x,q_x)$ that gets mapped into $[0,\langle f(x),0\rangle]$ by $s$.

If $x,y\in C_0$ and $f(x)<f(y)$ then $y\in(p_y,q_y)\setminus (p_x,q_x)$, and we conclude that $x\mapsto(p_x,q_x)$ is injective and $C_0$ is countable. Likewise $C_1$ is countable.

As above: let $x\in C(f)\cap A$ and assume $f(x)\in V$. Then $s(x)=\langle f(x),0\rangle$ or $s(x)=\langle f(x),1\rangle$. As above we split $C(f)\cap A$ into two pieces: $C_0=\{x:s(x)=\langle f(x),0\rangle\}$ and  $C_1=\{x:s(x)=\langle f(x),1\rangle\}$.

If $x\in C_0$ then we take a rational interval $(p_x,q_x)$ that gets mapped into $[0,\langle f(x),0\rangle]$ by $s$. If $x,y\in C_0$ and $f(x)<f(y)$ then $y\in(p_y,q_y)\setminus (p_x,q_x)$, and we conclude that $x\mapsto(p_x,q_x)$ is injective and $C_0$ is countable. Likewise $C_1$ is countable.