The "special case in $\mathbb N$" is a theorem of ZF:

Theorem (ZF). For any well-orderable set $X$ and any fixpoint-free function $f:X\to X$ there is a family $\mathfrak S\subseteq\mathcal P(X)$ such that $|\mathfrak S|\le3$ and $\bigcup\mathfrak S=X$ and $S\cap f(S)=\varnothing$ for all $S\in\mathfrak S$.

Proof. I will show that the graph $G=(X,E)$ with vertex set $X$ and edge set $E=\{\{x,f(x)\}:x\in X\}$ is $3$-colorable. Let $\{C_i:i\in I\}$ be the set of all connected components of $G$. (Of course each $C_i$ is $3$-colorable, being a unicyclic connected graph.) Fix a well-ordering of $X$, and let $x_i$ be the least element of $C_i$.

If $C_i$ contains no odd cycle, there is a unique proper red-blue vertex coloring of $C_i$ such that $x_i$ is red.

If $C_i$ contains a (necessarily unique) odd cycle, Let $y_i$ be the vertex on the cycle nearest to $x_i$. Then there is a unique proper red-blue-green coloring of $C_i$ such that $x_i$ is the only green vertex and $f(x_i)$ is red.

I don't know about your general statement, but it may be worth mentioning that it's not provable in ZF if we strengthen it by requiring the cardinal $\kappa_0$ to be an aleph:

Theorem. (ZF) Suppose that for any set $X$ and any fixpoint-free involution $f:X\to X$ there is a well-orderable family $\mathfrak S\subseteq\mathcal P(X)$ such that $\bigcup\mathfrak S=X$ and $S\cap f(S)=\varnothing$ for all $S\in\mathfrak S$. Then every collection of $2$-element sets has a choice function.

Proof. We may assume that the $2$-element sets are disjoint, and then they are the orbits of an involution.

The "special case in $\mathbb N$" is a theorem of ZF:

Theorem (ZF). For any well-orderable set $X$ and any fixpoint-free function $f:X\to X$ there is a family $\mathfrak S\subseteq\mathcal P(X)$ such that $|\mathfrak S|\le3$ and $\bigcup\mathfrak S=X$ and $S\cap f(S)=\varnothing$ for all $S\in\mathfrak S$.

Proof. I will show that the graph $G=(X,E)$ with vertex set $X$ and edge set $E=\{\{x,f(x)\}:x\in X\}$ is $3$-colorable. Let $\{C_i:i\in I\}$ be the set of all connected components of $G$. (Of course each $C_i$ is $3$-colorable, being a unicyclic connected graph.) Fix a well-ordering of $X$.

If $C_i$ contains no odd cycle, let $x_i$ be the least vertex in $C_i$. There is a unique proper red-blue vertex coloring of $C_i$ such that $x_i$ is red.

If $C_i$ contains a (necessarily unique) odd cycle, Let $x_i$ be the least vertex in $C_i$ which is not a cut vertex. Then there is a unique proper red-blue-green coloring of $C_i$ such that $x_i$ is the only green vertex and $f(x_i)$ is red.

I don't know about your general statement, but it may be worth mentioning that it's not provable in ZF if we strengthen it by requiring the cardinal $\kappa_0$ to be an aleph:

Theorem. (ZF) Suppose that for any set $X$ and any fixpoint-free involution $f:X\to X$ there is a well-orderable family $\mathfrak S\subseteq\mathcal P(X)$ such that $\bigcup\mathfrak S=X$ and $S\cap f(S)=\varnothing$ for all $S\in\mathfrak S$. Then every collection of $2$-element sets has a choice function.

Proof. We may assume that the $2$-element sets are disjoint, and then they are the orbits of an involution.