Recall that $\mathbf{non}(\mathcal{B})$ is the least cardinality of a non-meager subset of $\mathbb{R}$; the choice of presentation of $\mathbb{R}$ does not matter for this, so we take $\mathbb{R} = \mathcal{P}(\omega)$. It is well-known that consistently, $\mathbf{non}(\mathcal{B}) < \mathfrak{c}$; indeed, start from a model of CH and add $\aleph_2$-many Cohen reals.
Theorem. $\kappa \leq \mathbf{non}(\mathcal{B})$.
Proof. Suppose $\mathcal{F} \subseteq \mathcal{P}(\omega)$ is nonmeager. It suffices to show that $\mathcal{F}$ works for the definition of $\kappa$. Let $A, B$ be disjoint infinite subsets of $\omega$. Let $X$ be the set of all $F \subseteq \omega$ such that there are infinitely many $n \in A \cap F$ with $f(n) \not \in F$. I claim that $X$ is comeager. Indeed, for each finite $J \subseteq A$, let $X_J$ be the set of all $F \subseteq \omega$ such that there is some $n \in A \backslash J$ with $n \in A \cap F$ and $f(n) \not \in F$. Each $X_J$ is open dense, and $X = \bigcap_J X_J$.
Thus, consistently $\kappa < \mathfrak{c}$.
Incidentally, the following is perhaps a cleaner formulation of $\kappa$:
Claim. Let $\lambda$ be the least cardinality of a family $\mathcal{G}$ of subsets of $\omega$, such that for each increasing sequence $(x_n: n < \omega)$, there is some $F \in \mathcal{G}$ such that for infinitely many even $n < \omega$, $x_{n} \in F$ but $x_{n+1} \not \in F$. Then $\kappa = \lambda$.
Proof. It is straightforward to check that $\lambda \leq \kappa$: given $\mathcal{F}$ witnessing the definition of $\kappa$ and $(x_n: n < \omega)$, write $A = \{x_{2n}: n < \omega\}$, $B = \{x_{2n+1}: n < \omega\}$ and let $f: A \to B$ be $x_{2n} \mapsto x_{2n+1}$.
So we show $\kappa \leq \lambda$. Let $\mathcal{G}$ witness the definition of $\lambda$; let $\mathcal{F} = \mathcal{G} \cup \{\omega \backslash S: S \in \mathcal{G}\}$. We show that $\mathcal{F}$ is as in the definition of $\kappa$.
Suppose $A, B$ are infinite disjoint sets and $f: A \to B$ is a bijection. Note that since $\mathcal{F}$ is closed under complements, we are free to replace $(A, B, f)$ by $(B, A, f^{-1})$, if desired.
Choose $A' \subseteq A$ infinite such that either for all $x \in A'$, $f(x) > x$, or else for all $x \in A'$, $f(x) < x$. After possibly interchanging $A$ and $B$, we can suppose that for all $x \in A'$, $f(x) > x$. Write $B' = f[A']$.
Define an increasing sequence $(x_n: n < \omega)$ inductively so that for all $n$, $x_{2n} \in A'$ and $x_{2n+1} = f(x_{2n}) \in B'$. Namely, let $x_0 = \min(A')$, let $x_1 = f(x_0)$, and having defined $x_{2n-1}$, let $x_{2n}$ be the least element of $A'$ bigger than $x_{2n-1}$, and let $x_{2n+1} = f(x_{2n})$.
Choose $F \in \mathcal{F}$ such that there are infinitely many even $n < \omega$ with $x_n \in F$ but $x_{n+1} \not \in F$. Then for each such $n$, $x_n \in A \cap F$ but $f(x_n) = x_{n+1} \not \in F$, so $F$ is as desired.
Remark. For each infinite $I \subseteq \omega \times \omega$, we get a variant notion $\kappa_I$ of $\kappa$, namely $\kappa_I$ is the least cardinality of a family $\mathcal{G}$ of subsets of $\omega$, such that for each increasing sequence $(x_n: n < \omega)$, there is some $F \in \mathcal{G}$ such that for infinitely many $(n, m) \in I$, $x_n \in F$ but $x_m \not \in F$ (possibly $\kappa_I = \infty$).
It is easy to check that $\kappa = \kappa_{\{(2n, 2n+1): n < \omega\}}$ and $\mathfrak{s} = \kappa_{\{(n, n+1): n < \omega\}}$. The above proof shows that each $\kappa_I \leq \mathbf{non}(\mathcal{B})$, provided that for each finite $J \subset \omega$, there are infinitely many $(n, m) \in I$ with $n, m \not \in J$. It is interesting to ask which of these $\kappa_I$'s can be separated...
