Quantum fluctuations in the kink sectors of Sin-Gordon and the quartic interaction theory are described by reflectionless Pöschl-Teller-Operators, which form a SUSY-Chain with $N$ elements. The second quantization is then obtained by the spectral decompostion of those operators that are connected through annihilation and creation operators. Sin Gordon corresponds to $N=1$ and $\Phi^4-$Theory to $N=2$. Because of the similarities of the quantum structure of both systems, there should be the possibility to construct a fermion model which corresponds to the quartic theory for certain values of the coulping constant, too. An interesting point is, that the eigenfunctions of the creation operators for the Nth system have the same structure as the $N$ body partition function in the Ising-Model, calculated in mean field approximation.
Edit: In short form: If you expand around the kink the second order term is called the fluctuation operator
$$\frac{\delta^2S_E}{\delta\phi(x)\delta\phi(x')}|_{\phi_{kink}=\phi}=[-\partial^2_x+V''(\phi_{kink}(x))]\delta(x-x')$$
The operators for both systems are of the form
$$F=-\partial^2_x+N^2m^2-N(N+1)m^2sech^2(mx)$$. With spectra
$$spec(F)=\{\omega^2_n\}_{n=0,...,l-1}\cup\{\omega^2_l\}_{l=N}\cup\{k^2+N^2m^2\}_{k\in\mathbb{R} }$$
This is a chain of supersymmetric operators. Another motivation for SUSY is this: The kink forms an equipotential surface in field configuration space due to its translational invariance. That means there is a parameter, where under change of it the action remains constant. So the fluctuations vanish. From this fact it follows, that the zero mode of the operator $F$ has to be the derivative along that parameter of the classical solution, which is a monotonous function, so the derivative has no nods and therefore it has to be the ground state of a supersymetrical quantum system. If you know the ground state of such a system you can evaluate the superpotential together with the ladder operators and recover the fluctuation operator from $F=A^\dagger_NA_N$.
$$A^\dagger_N=-\partial_x+Nmtanh(mx)$$
$$A_N=\partial_x+Nmtanh(mx)$$
The Nth system has $N-1$ bound states which are all smooth and faster decaying torwards infinity than any polynomial.
Sin-Gordon (normalized):
$\omega^2_0=0$ $$\psi_0(x)=\sqrt{\frac{m}{2}}sech(mx)$$
$\phi^4$ (normalized):
$\omega^2_0=0$ $$\psi_0(x)=\sqrt{\frac{3m}{4}}sech^2(mx)$$
$\omega^2_1=3m^2$
$$\psi_1(x)=\sqrt{\frac{3m}{2}}tanh(mx)sech(mx)$$
And the normalized and generalized eigenfunctions for the continious part of the spectra respectively
$k\geq0$
$$Y_k(x)=\frac{(tanh(mx)-ik)e^{ikx}}{\sqrt{k^2+m^2}}e^{i\delta_k}$$
$k<0$
$$Y_k(x)=\frac{(tanh(mx)-ik)e^{ikx}}{\sqrt{k^2+m^2}}e^{-i\delta_k}$$
and
$k\geq0$
$$Y_k(x)=\frac{(-k^2-3imktanh(mx)+4m^2+6m^2sech^2(mx))e^{ikx}}{\sqrt{(k^2+m^2)(k^2+4m^2)}}e^{i\delta_k}$$
$k<0$
$$Y_k(x)=\frac{(-k^2-3imktanh(mx)+4m^2+6m^2sech^2(mx))e^{ikx}}{\sqrt{(k^2+m^2)(k^2+4m^2)}}e^{-i\delta_k}$$
where $\delta_k$ are the phase factors. These functions can easely be found under use of the ladder structure and an ansatz of plane waves. They form a set of generalized eigenfunction under construction of a gelfand triple. The second quantization for the vacuum sectors is now straight forward as in the free case, but the kink sector is a bit tricky in terms of fock space construction and introduction of a distorted fourier transform associated with the generalized eigenfunctions. It should be also clear, that the zero modes cause somee trouble regarding divergencies of the hamiltionian. They have to be eliminated by normal ordering with respect to the kink what is called kink representation and there exists a unitary map between kink and vacuum representation which involves the semiclassical 1-loop mass correction. Basicly the fields and the conjugated momenta in heisenberg representation associated to the SUSY-Chain look like this after a lot of struggle.
$$\phi(x,t)=-\sqrt{M_{cl}}\mathcal{X}\psi_0+\displaystyle\sum_{n=1}^{N-1} \frac{1}{\sqrt{\omega_n}}(a_n(t)+a^{\dagger}_n(t))\psi_n+\frac{1}{\sqrt{2\pi}}\int\mathrm {\frac{1}{\sqrt{\omega_k}}}(a_k(t)Y_k(x)+a^{\dagger}_k(t)\overline{Y_k(x)})$$
$$\pi(x,t)=-\frac{\mathcal{P}}{\sqrt{M_{cl}}}\psi_0-\displaystyle\sum_{n=1}^{N-1} \frac{i\sqrt{\omega_n}}{\sqrt{2}}(a_n(t)+a^{\dagger}_n(t))\psi_n+\frac{1}{\sqrt{2\pi}}\int\mathrm {\frac{-i\sqrt{\omega_k}}{\sqrt{2}}}(a_k(t)Y_k(x)-a^{\dagger}_k(t)\overline{Y_k(x)})$$
These satisfy the CCR for all schwartz functions and it can be shown that there exist fock spaces on which the associated hamiltonians to the systems are self adjoint operators. I have to read the paper of Coleman again but it should be possible in priniple to construct a dual fermion system to any of the elements of the chain like it was shown by coleman for Sin-Gordon in the 70s.
Edit: I have to slow down my enthusiasm a bit because there exists no multikink configuration in $\phi^4$ as it is present in Sin-Gordon. Maybe one should concentrate on the dilute kink gas in $\phi^4$ if someone searches for a fermion field configurations dual to the scalar field configurations. But that fermion model cannot be a multiparticle system if you want it to have the same duality like the massive thirring model and the sin gordon model have, because there are no solutions with topological charge greater than $\pm 1$ due to the vacuum structure and the charge is associated with the fermion number. If a duality exists it has to be of completely different nature.
Best Answer
In general, the Wetterich equation reads $$k \partial_k \Gamma_k = \frac{1}{2} {\rm Tr}\left[\left(\Gamma^{(2)}_k + R_k \right)^{-1} \cdot k \partial_k R_k \right],$$ where $\Gamma^{(2)}_k$ is a matrix of second-order derivatives of $\Gamma^{(2)}_k$ with respect to its arguments (here, $\phi$, $\psi$, and $\bar{\psi}$) and $R_k$ is the corresponding matrix of regulators. This is Eq. (4) in the paper you linked.
The original expression for the Wetterich equation missed an inverse on $\left(\Gamma^{(2)}_k + R_k \right)^{-1}$ before the edit. This inverse is important in general, because even if the regulator $R_k$ only couples to $\phi \phi$ and $\psi \bar{\psi}$ terms, the inverse involves all $9$ derivatives of $\Gamma_k$, not just $\delta^2 \Gamma_k/\delta \phi \delta \phi$ and $\delta^2 \Gamma_k/\delta \psi \delta \bar{\psi}$. This means that, in order to evaluate this inverse, in principle you will need to make your ansatz for $\Gamma_k$, evaluate all $9$ second derivatives, and then set $\phi$, $\psi$, and $\bar{\psi}$ to constant values, independent of space/momenta, so that you can actually evaluate the inverse. (The cited paper does not mention this point explicitly, but since they include a local potential $U_k(\phi^2/2)$ in their ansatz, they presumably take $\phi$ to be a scalar variable, and perhaps they set the $\psi$'s to be either zero or some other constant value). In this particular example, the inverse does simplify nicely when evaluated at zero fields (as shown for a simplified example below); the issue then comes from the fact that one needs to define the running fields in terms of derivatives of $\Gamma_k$ and differentiate the flow equation before evaluating at zero fields.
Note that the ansatz for $\Gamma_k$ that you have written didn't denote which quantities were assumde to run with $k$ (in the original post, before editing), so just to compare to the cited paper's ansatz (Eq. (5)), $$\Gamma_k[\phi,\psi,\bar{\psi}] = \int dx~\left[ Z_k (\partial_\mu \phi)^2 + U_k(\phi^2/2) + Z_{\psi,k} \bar{\psi} i \gamma^\mu \partial_\mu \psi + i h_k \phi \bar{\psi} \psi \right].$$ It appears that you wish to explicitly truncate the authors' $U_k(\phi^2/2)$ at second order in $\phi^2$ (around $\phi=0$) and introduce an $i \bar{\psi} M_k \psi$. You will need to define how each coefficient is extracted from $\Gamma_k$ in order to compute your flow equations. For example, by defining $m^2_k = \lim_{p \rightarrow 0}\frac{\delta^2 \Gamma_k}{\delta \phi(p) \delta \phi(q)}\Big|_{\phi(p) \rightarrow \phi,\psi \rightarrow 0,\bar{\psi} \rightarrow 0}$, $M_k = \lim_{p \rightarrow 0}\frac{1}{i} \frac{\delta^2 \Gamma_k}{\delta \bar{\psi}(p) \delta \psi(q)}\Big|_{\phi(p) \rightarrow \phi,\psi \rightarrow 0,\bar{\psi} \rightarrow 0}$, $Z_k = \lim_{p \rightarrow 0}\partial_{p^2}\left[\frac{\delta^2 \Gamma_k}{\delta \phi(p) \delta \phi(-p)}\Big|_{\phi(p) \rightarrow \phi,\psi \rightarrow 0,\bar{\psi} \rightarrow 0}\right]$, etc. This will involve taking derivatives of the flow equation with respect to the fields in order to get $k \partial_k m^2_k$, $k \partial_k M_k$, etc. In doing so, it is again important to treat the inverse $\left(\Gamma^{(2)}_k + R_k \right)^{-1}$ carefully, since this will generate non-trivial terms in the flow equations for these quantities (using the identity $\frac{\delta}{\delta \phi} A^{-1} = -A^{-1} \frac{\delta A}{\delta \phi} A^{-1}$). See, for example, this paper for an example application to $\phi^4$ theory.
Edit: When multiple fields are involved, as in this case, or in a non-equilibrium problem, the inverse is a matrix inverse in the fields (when converted to momentum space). I'll sketch it for the cited paper's ansatz simplifed to $U_k(\phi^2/2) = m_k^2 \phi(x)^2/2$. In momentum space (assuming no mistakes...) $$\begin{array}{c c}\Gamma_k &= \int dp_1 dp_2~\Big[(Z_k p^2 + m^2_k) \phi(p_1)\phi(p_2)\delta(p_1 + p_2) + Z_{\psi,k} i \bar{\psi}(p_1) \gamma^\mu (p_1)_\mu \psi(p_2) \delta(p_1+p_2)\Big] \\ & ~~~~~~~~ + \int dp_1 dp_2 dp_3~ i h_k \phi(p_1)\bar{\psi}(p_2) \psi(p_3) \delta(p_1 + p_2 + p_3)\end{array}.$$ (I am not very familiar with dealing with fermionic fields, so you'll want to double-check my math here).
Then, the matrix of derivatives is $$\Gamma^{(2)}_k = \begin{bmatrix} \frac{\delta \Gamma_k}{\delta \phi(p) \delta \phi(q)} & \frac{\delta \Gamma_k}{\delta \phi(p) \delta \psi(q)} & \frac{\delta \Gamma_k}{\delta \phi(p) \delta \bar{\psi}(q)} \\ \frac{\delta \Gamma_k}{\delta \psi(p) \delta \phi(q)} & \frac{\delta \Gamma_k}{\delta \psi(p) \delta \psi(q)} & \frac{\delta \Gamma_k}{\delta \psi(p) \delta \bar{\psi}(q)} \\ \frac{\delta \Gamma_k}{\delta \bar{\psi}(p) \delta \phi(q)} & \frac{\delta \Gamma_k}{\delta \bar{\psi}(p) \delta \psi(q)} & \frac{\delta \Gamma_k}{\delta \bar{\psi}(p) \delta \bar{\psi}(q)} \end{bmatrix} = \begin{bmatrix} (Z_k p^2 + m_k^2)\delta(p+q) & \int dp_2~ i h_k \bar{\psi}(p_2) \delta(p+q+p_2) & \int dp_3~ i h_k \psi(p_3) \delta(p+q+p_3) \\ \int dp_2~ i h_k \bar{\psi}(p_2) \delta(p+q+p_2) & 0 & Z_{k,\psi} i \gamma^\mu p_\mu \delta(p+q) + \int dp_1~ i h_k \phi(p_1) \delta(p+q+p_1) \\ \int dp_3~ i h_k \psi(p_3) \delta(p+q+p_3) & Z_{k,\psi} i \gamma^\mu p_\mu \delta(p+q) + \int dp_1~ i h_k \phi(p_1) \delta(p+q+p_1) & 0 \end{bmatrix}$$
Then, one would add to this the matrix $$R_k = \begin{bmatrix} R_{\phi,k} & 0 & 0 \\ 0 & 0 & R_{\psi,k} \\ 0 & R_{\psi,k} & 0 \end{bmatrix}\delta(p+q)$$
To calculate the inverse in practice, we need to choose functions $\phi(p)$, $\psi(p)$, and $\bar{\psi}(p)$ at which we wish to evaluate the inverse, since we cannot do it for arbitrary functions. Typically, one chooses momentum-independent values. If we had kept the full $\phi$ dependence of $U_k(\phi^2/2)$, we would typically set $\phi(p) = \phi$ (a scalar), but since we've truncated it, we could set all of $\phi(p) = \psi(p) = \bar{\psi}(p) = 0$, giving $$\Gamma^{(2)}_k + R_k = \begin{bmatrix} (Z_k p^2 + m_k^2) + R_{\phi,k} & 0 & 0 \\ 0 & 0 & Z_{k,\psi} i \gamma^\mu p_\mu + R_{\psi,k} \\ 0 & Z_{k,\psi} i \gamma^\mu p_\mu + R_{\psi,k} & 0 \end{bmatrix}\delta(p+q),$$ which can now be evaluated as a matrix inverse (the inverse of the momentum-dependence here is just another $\delta(p+q)$). For this particular choice, the inverse does end up just being 1 over the non-zero elements. If we had chosen constant values $\phi = \phi_0$, $\psi = \psi_0$, and $\bar{\psi} = \bar{\psi}_0$, the inverse would not be as trivial.
Now, for the choice $\phi(p) = \psi(p) = \bar{\psi}(p) = 0$, you'll notice that the dependence on $h_k$ dropped out when we set the fields to $0$. That does not mean $h_k$ would not flow in this model because, as I mentioned above, one needs to define $h_k$ in terms of derivatives of $\Gamma_k$, e.g., $h_k = \frac{1}{i}\frac{\delta^3 \Gamma_k}{\delta \phi(p_1) \delta \bar{\psi}(p_2) \delta \psi(p_3)}\Big|_{\phi = \bar{\psi} =\psi = 0}$, and then differentiate the flow equation with respect to these fields before setting $\phi = \bar{\psi} =\psi = 0$. The result should be a non-trivial flow equation for $h_k$ (coupled to other running parameters).