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OK, third time's the charm, hopefully:

First, a lemma:

Fix an arbitrary successor ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d_f$ with the same domain such that $f\le d_f$ and $d_f$ takes on only finitely many values.

The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor-of-a-successor case.

Now suppose the proof holds for all non-limit $\gamma<\lambda+1$ for $\lambda$ a limit, and take $\beta=\lambda+1$. Fix some successor ordinal $\chi$ with $\chi<\lambda$ and $\chi>f(\lambda)$; such a $\chi$ must exist since $\lambda$ is limit and $f$ is regressive.

But now consider the map $F=f\upharpoonright \chi$. Since $\lambda$ is limit and $\chi<\lambda$ is a successor ordinal, we may apply the induction hypothesis to get a $d_{F}$. But now consider the map $d_f$ with domain $\lambda-\{0\}$ extending $d_{F}$ and taking on the value $max\{f(\lambda), d_F(\epsilon)\}$ (where $\epsilon+1=\chi$) on all inputs $\theta$ with $\chi<\theta \le\lambda$. This function has the desired properties - in particular, it is regressive by choice of $\chi$ - so we are done. QED

Now, to answer the actual question: fix a nondecreasing regressive $f$ defined on $\omega_1-\{0\}$. By Fodor's lemma + nondecreasingness, $f$ is eventually constant; fix some countable successor ordinal $\alpha$ such that for all $\beta>\alpha$, $f(\beta)=f(\alpha)=\gamma$.

Apply the lemma to $f\upharpoonright \alpha$ to get a function $d_0$; now extend $d_0$ to a function $d$ defined on all of $\omega_1-\{0\}$ by setting $d(\theta)=max\{f(\alpha), d_0(\epsilon)\}$ (where $\epsilon+1=\alpha$) for all $\theta\ge\alpha$.

Edit. I accepted, unaccepted, and again accepted the answer: It is incomplete (as comments below show), so the question is open, but I got enough from it, just to clarify to myself what I was asking, I'll keep thinking about it. Thank you

OK, third time's the charm, hopefully:

First, a lemma:

Fix an arbitrary successor ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d_f$ with the same domain such that $f\le d_f$ and $d_f$ takes on only finitely many values.

The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor-of-a-successor case.

Now suppose the proof holds for all non-limit $\gamma<\lambda+1$ for $\lambda$ a limit, and take $\beta=\lambda+1$. Fix some successor ordinal $\chi$ with $\chi<\lambda$ and $\chi>f(\lambda)$; such a $\chi$ must exist since $\lambda$ is limit and $f$ is regressive.

But now consider the map $F=f\upharpoonright \chi$. Since $\lambda$ is limit and $\chi<\lambda$ is a successor ordinal, we may apply the induction hypothesis to get a $d_{F}$. But now consider the map $d_f$ with domain $\lambda-\{0\}$ extending $d_{F}$ and taking on the value $max\{f(\lambda), d_F(\epsilon)\}$ (where $\epsilon+1=\chi$) on all inputs $\theta$ with $\chi<\theta \le\lambda$. This function has the desired properties - in particular, it is regressive by choice of $\chi$ - so we are done. QED

Now, to answer the actual question: fix a nondecreasing regressive $f$ defined on $\omega_1-\{0\}$. By Fodor's lemma + nondecreasingness, $f$ is eventually constant; fix some countable successor ordinal $\alpha$ such that for all $\beta>\alpha$, $f(\beta)=f(\alpha)=\gamma$.

Apply the lemma to $f\upharpoonright \alpha$ to get a function $d_0$; now extend $d_0$ to a function $d$ defined on all of $\omega_1-\{0\}$ by setting $d(\theta)=max\{f(\alpha), d_0(\epsilon)\}$ (where $\epsilon+1=\alpha$) for all $\theta\ge\alpha$.

OK, third time's the charm, hopefully:

First, a lemma:

Fix an arbitrary successor ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d_f$ with the same domain such that $f\le d_f$ and $d_f$ takes on only finitely many values.

The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor-of-a-successor case.

Now suppose the proof holds for all non-limit $\gamma<\lambda+1$ for $\lambda$ a limit, and take $\beta=\lambda+1$. Fix some successor ordinal $\chi$ with $\chi<\lambda$ and $\chi>f(\lambda)$; such a $\chi$ must exist since $\lambda$ is limit and $f$ is regressive.

But now consider the map $F=f\upharpoonright \chi$. Since $\lambda$ is limit and $\chi<\lambda$ is a successor ordinal, we may apply the induction hypothesis to get a $d_{F}$. But now consider the map $d_f$ with domain $\lambda-\{0\}$ extending $d_{F}$ and taking on the value $max\{f(\lambda), d_F(\epsilon)\}$ (where $\epsilon+1=\chi$) on all inputs $\theta$ with $\chi<\theta \le\lambda$. This function has the desired properties - in particular, it is regressive by choice of $\chi$ - so we are done. QED

Now, to answer the actual question: fix a nondecreasing regressive $f$ defined on $\omega_1-\{0\}$. By Fodor's lemma + nondecreasingness, $f$ is eventually constant; fix some countable successor ordinal $\alpha$ such that for all $\beta>\alpha$, $f(\beta)=f(\alpha)=\gamma$.

Apply the lemma to $f\upharpoonright \alpha$ to get a function $d_0$; now extend $d_0$ to a function $d$ defined on all of $\omega_1-\{0\}$ by setting $d(\theta)=max\{f(\alpha), d_0(\epsilon)\}$ (where $\epsilon+1=\alpha$) for all $\theta\ge\alpha$.

OK, third time's the charm, hopefully:

First, a lemma:

Fix an arbitrary successor ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d_f$ with the same domain such that $f\le d_f$ and $d_f$ takes on only finitely many values.

The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor-of-a-successor case.

Now suppose the proof holds for all non-limit $\gamma<\lambda+1$ for $\lambda$ a limit, and take $\beta=\lambda+1$. Fix some successor ordinal $\chi$ with $\chi<\lambda$ and $\chi>f(\lambda)$; such a $\chi$ must exist since $\lambda$ is limit and $f$ is regressive.

But now consider the map $F=f\upharpoonright \chi$. Since $\lambda$ is limit and $\chi<\lambda$ is a successor ordinal, we may apply the induction hypothesis to get a $d_{F}$. But now consider the map $d_f$ with domain $\lambda-\{0\}$ extending $d_{F}$ and taking on the value $max\{f(\lambda), d_F(\epsilon)\}$ (where $\epsilon+1=\chi$) on all inputs $\theta$ with $\chi<\theta \le\lambda$. This function has the desired properties - in particular, it is regressive by choice of $\chi$ - so we are done. QED

Now, to answer the actual question: fix a nondecreasing regressive $f$ defined on $\omega_1-\{0\}$. By Fodor's lemma + nondecreasingness, $f$ is eventually constant; fix some countable successor ordinal $\alpha$ such that for all $\beta>\alpha$, $f(\beta)=f(\alpha)=\gamma$.

Apply the lemma to $f\upharpoonright \alpha$ to get a function $d_0$; now extend $d_0$ to a function $d$ defined on all of $\omega_1-\{0\}$ by setting $d(\theta)=max\{f(\alpha), d_0(\epsilon)\}$ (where $\epsilon+1=\alpha$) for all $\theta\ge\alpha$.

Edit. I accepted, unaccepted, and again accepted the answer: It is incomplete (as comments below show), so the question is open, but I got enough from it, just to clarify to myself what I was asking, I'll keep thinking about it. Thank you

Something much stronger is true:

Fix an arbitrary ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d_f$ with the same domain such that $f\le d_f$ and $d_f$ takes on only finitely many values.

The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor case.

Now suppose the proof holds for all $\gamma<\lambda$ for $\lambda$ a limit, and take $\beta=\lambda$. Since $f$ is regressive and nondecreasing and $\lambda$ is a limit, there must be some $\gamma<\lambda$ such that $\forall \gamma<\eta<\lambda$, we have $f(\gamma)=f(\eta)$.

But now consider the map $F=f\upharpoonright \gamma+1. $ Since $\lambda$ is limit and $\gamma<\lambda$, $\gamma+1$ is also $<\lambda$, so we may apply the induction hypothesis to get a $d_{F}$. But now consider the map $d_f$ with domain $\lambda-\{0\}$ extending $d_{F}$ and taking on the value $d_{F}(\gamma)$ on all inputs $>\gamma$. This function has the desired properties, by assumption on $\gamma$.

OK, third time's the charm, hopefully:

First, a lemma:

Fix an arbitrary successor ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d_f$ with the same domain such that $f\le d_f$ and $d_f$ takes on only finitely many values.

The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor-of-a-successor case.

Now suppose the proof holds for all non-limit $\gamma<\lambda+1$ for $\lambda$ a limit, and take $\beta=\lambda+1$. Fix some successor ordinal $\chi$ with $\chi<\lambda$ and $\chi>f(\lambda)$; such a $\chi$ must exist since $\lambda$ is limit and $f$ is regressive.

But now consider the map $F=f\upharpoonright \chi$. Since $\lambda$ is limit and $\chi<\lambda$ is a successor ordinal, we may apply the induction hypothesis to get a $d_{F}$. But now consider the map $d_f$ with domain $\lambda-\{0\}$ extending $d_{F}$ and taking on the value $max\{f(\lambda), d_F(\epsilon)\}$ (where $\epsilon+1=\chi$) on all inputs $\theta$ with $\chi<\theta \le\lambda$. This function has the desired properties - in particular, it is regressive by choice of $\chi$ - so we are done. QED

Now, to answer the actual question: fix a nondecreasing regressive $f$ defined on $\omega_1-\{0\}$. By Fodor's lemma + nondecreasingness, $f$ is eventually constant; fix some countable successor ordinal $\alpha$ such that for all $\beta>\alpha$, $f(\beta)=f(\alpha)=\gamma$.

Apply the lemma to $f\upharpoonright \alpha$ to get a function $d_0$; now extend $d_0$ to a function $d$ defined on all of $\omega_1-\{0\}$ by setting $d(\theta)=max\{f(\alpha), d_0(\epsilon)\}$ (where $\epsilon+1=\alpha$) for all $\theta\ge\alpha$.