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I have been thinking about this question since it was bumped up. I am burned out now, so I thought I would post a few basic things that I noted.

First, as you noted that any function $f \in \mathcal F$ would reach a constant value after some point. Also that value will obviously be the maximum value of the function. So suppose that $f:\omega_1 \rightarrow \omega_1$ maximises at a value $v_f$. Now it seems to me that this variable $v_f$ may be of some significance for the given question.

For example, let $\mathcal F_\alpha \subset \mathcal F$ ($\alpha<\omega_1$) denote the collection of functions $f \in \mathcal F$, which satisfy the additional property the $v_f < \alpha$. Your original question was (A) Giving a function $\psi : \mathcal F \to \mathcal K$ (satisfying certain properties of course). Now I think we can also consider a sub-question: (B) Can one give a function $\psi_\alpha : \mathcal F_\alpha \to \mathcal K$ for any arbitrary $\alpha < \omega_1$ ($\psi_\alpha$ and $\mathcal K$ satisfying properties as described in question).

For example, here is an example of a function $\psi_{\omega^2} : \mathcal F_{\omega^2} \to \mathcal K$. I think, I can give a similar description for $\psi_{\omega^3} : \mathcal F_{\omega^3} \to \mathcal K$. I have not added it for the sake of brevity, but if you think the example for $\psi_{\omega^2}$ is not illustrative enough, then I will add it.


An example for $\psi_{\omega^2}$. Given a function $f \in \mathcal F_{\omega^2}$, first determine the maximum value $v_f$ of the function. Now suppose that the maximum value $v_f$ is of the form $\omega \cdot a+b$ (with $a \in \mathbb{N}^+$, $b \in \mathbb{N}$). Denote $\psi_{\omega^2} (f)$ as $F$. Now define $F$ as follows: $$F(x)=\omega \cdot a+b=v_f \qquad \mathrm{for} \quad x > \omega \cdot(a+1)$$ $$F(x)=\omega \cdot a \qquad \mathrm{for} \quad \omega\cdot a < x \leq \omega \cdot (a +1)$$ $$F(x)=\omega \cdot n \qquad \mathrm{for} \quad \omega\cdot n < x \leq \omega \cdot (n+1)$$ $$F(x)=f(x) \qquad \mathrm{for} \quad x \leq \omega $$ In the third line, we have $1 \leq n < a$. (EDIT : Made a correction in the first and second equation)

One line of thinking is to see whether we can keep giving the function $\psi_\alpha$ or not. If no, then what is the point at which we can no longer do that (the answer to (A), the question in OP, would be negative in that case). If yes, in that case the answer to (B) would be positive. Also, in that case, what would be the general requirements/obstacles at each level (to determine whether (B) implies (A) or not). Anyway, just a suggestion that might perhaps be useful (and, to be fair, possibly not useful).

EDIT2:

Regarding the description of $\psi_{\omega^2} (f)=F$, it was rightly mentioned in comments that it is only guaranteed to work for those functions $f \in \mathcal F_{\omega^2}$ which increase smoothly. In other words, if $v_f$ is the maximum value of $f$, then $f$ satisfies the condition $\mathrm{range}(f)=v_f+1$.

Somewhat briefly, I just wanted to point out how this can be rectified somewhat easily (though it seems it would be at the cost of doubling the number of equations above). Suppose we had some function $f \in \mathcal F_{\omega^2}$ so that $f(\omega \cdot 2)=\omega+2$ and $f(\omega \cdot 3)=\omega.2+4$. Further suppose the max. value for $f$ is: $v_f=f(\omega \cdot 3)$.

Now we define $\psi_{\omega^2} (f)=F$ as follows:

(1) For $x \leq \omega$, set $F(x)=f(x)$

(2) Set $F(\omega+1)=\omega$, $F(\omega+2)=\omega+1$

(3) For $\omega+3 \leq x \leq \omega \cdot 2$, set $F(x)=f(\omega \cdot 2)$

(4) Set $F(\omega \cdot 2+1)=\omega \cdot 2$, $F(\omega \cdot 2+2)=\omega \cdot 2+1$, $F(\omega \cdot 2+3)=\omega \cdot 2+2$, $F(\omega \cdot 2+4)=\omega \cdot 2+3$

(5) For $\omega \cdot 2+5 \leq x \leq \omega \cdot 3$, set $F(x)=f(\omega \cdot 3)$

(6) For $x > \omega \cdot 3$, set $F(x)=f(\omega \cdot 3)$

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