The reason you are getting the wrong Jacobian is that you are not landing into a <PERSON> measure at the end. The Jacobian that the paper writes is the one that exchange $\mathcal{D}A$ by the <PERSON> measure of $U$. So first you need to parametrize the gauge group like $U(\alpha)=e^{iT_{i}\alpha^{i}}$ and then write the <PERSON> measure in terms of $\alpha$, something like $d\mu(\alpha)$, and finally compute the Jacobian for $\mathcal{D}A\rightarrow d\mu(\alpha)$.
A faster way to obtain the Jacobian is by noting that the measure $$ (\det (\partial_z - A_z))^{-1}(\det (\partial_\bar z - A_\bar z))^{^{-1}}\mathcal{D}A $$ is invariant under gauge transformation $A_{m}\rightarrow A_{m}+[D_{m},\alpha(x)]$ (the signs might be wrong). If you restrict the path integral over $A$ to be just over pure gauge configurations you get a <PERSON> measure for $U$, since it satisfies all the properties of a <PERSON> measure. Recall that the <PERSON> measure is unique up to an overall scaling. So you can identify
$$ \mathcal{D}U=(\det (\partial_z - A_z))^{-1}(\det (\partial_\bar z - A_{\bar z})^{^{-1}}\mathcal{D}A $$
where $A$ is restricted to be pure gauge. 418
So we get the Jacobian:
$$ \mathcal{D}A=(\det (\partial_z - \partial_zUU^{-1}))(\det (\partial_\bar z - \partial_{\bar z}UU^{-1})\mathcal{D}U $$
You can motivate it in a more rigorous way by noting that you can use the Faddeev-Popov procedure in reverse to obtain the <PERSON> measure in terms of $A$. You define the Faddeev-Popov term by the relation
$$ 1=\Delta_{FP}(A_0)\int \mathcal{D}U \delta(A-A_0^U) $$
where $A_0^{U}$ is the gauge connection obtained by acting the action of $U$ on a fixed connection $A_0$. Inserting this $1$ on a path integral on $A$ we can use the delta to trade any functional of $A$ into a functional of $A_0^{U}$ by integrating on $A$ first. You are left with a path integral over $U$ with the correct Jacobian.
For the $\delta(\tilde{f})$ you can use the formula
$$ \prod_{i=1}^{n}\delta (O^{i}_jb^j)=(det(O))^{-1}\delta^{n}(b) $$
applied to the continuum, $n\rightarrow\infty$, $i\rightarrow x$ and $b_i\rightarrow f(x)$. There is just a small subtlety that you kindly remind me. This is the fact that since $(\varepsilon_{z},\varepsilon_{\bar z})$ is defined to be orthogonal to pure gauge configurations, the delta function $\delta(\tilde f)$ factorizes into one delta just involving $\varepsilon_{z}$-terms times another delta involving just the $\varepsilon_{\bar z}$-terms. 418