Propagation of a wavefunction on a Riemannian sigma model

I have a question about Riemannian sigma model, in particular how wavefunctions propagate. Here the Riemannian sigma model refers to the one introduced in 10.4.1 and 10.4.2 of the book $$\ulcorner$$ K. Hori et al., Mirror Symmetry, 2003$$\lrcorner$$. Let me write down the data.

• We work in an Euclidean time.
• It has a spacetime manifold $$M=S^1$$ and a Riemannian target $$(X,g)$$. On a target, we have a potential $$h:X\rightarrow \mathbb{R}$$.
• It has a bosonic field $$\phi:M\rightarrow X$$ and fermionic fields $$\psi,\overline{\psi}\in \Gamma(M,\phi^*TX)$$ whose degrees are shifted by $$-1$$ and $$1$$, respectively.
• The Hilbert space is a de Rham complex, i.e., $$\mathcal{H}=\Omega^{\bullet}(X)$$.
• Supercharges are presented as $$Q=d+dh\wedge,\quad \overline{Q}=Q^{\dagger}=d^{\dagger}+\iota_{\text{grad}h},\quad F=\text{degree of the form}, \quad H=\frac{1}{2}\{Q,\overline{Q}\}.$$

As mentioned in the beginning, I want to know how wavefunctions propagate. Consider a bosonic particle moving on a real line, with a potential $$V:\mathbb{R}\rightarrow\mathbb{R}$$. The Hilbert space if $$L^2(\mathbb{R})$$ and for $$t>t_0$$ we have $$\Psi(x,t)=\int_{\mathbb{R}}dx_0 K(x_0,t_0;x,t)\Psi(x_0,t_0)$$ where $$K(x_0,t_0;x,t)$$ has two interpretations as the path integral $$\int_{\phi(t_0)=x_0, \ \phi(t)=x} D\phi \ e^{-S_E[\phi]}$$ (here we set $$\hbar=1$$ and use Euclidean action $$S_E$$).

I want a similar expression for Riemannian sigma model. A naive try will be $$\Psi(x,t)=\int_{X}dx_0 K(x_0,t_0;x,t)\Psi(x_0,t_0) \quad \text{for} \quad K(x_0,t_0;x,t)=\int_{\phi(t_0)=x_0 \ \phi(t)=x} D\phi D\psi D\overline{\psi} e^{-S_E[\phi,\psi,\overline{\psi}]},$$ where when integrating over $$x_0$$ we use the volume form on $$X$$ determined by the metric $$g$$. However, it is a nonsense because $$\Psi(x,t)\in \wedge^{\bullet}T^*_x X$$ whereas $$\Psi(x_0,t_0)\in \wedge^{\bullet}T^*_{x_0}X$$. (Recall that $$\mathcal{H}=\Omega^{\bullet}(X)$$ so that wavefunctions are differential forms). To compare $$\Psi(x,t)$$ and $$\Psi(x_0,t_0)$$ we will need a connection $$\nabla$$ on the vector bundle $$\wedge^r T^*X$$ (if $$\Psi(-,t_0)$$ is an $$r$$-form). Therefore our second attempt will be $$\Psi(x,t)=\int_{X}dx_0 K(x_0,t_0;x,t)\Pi_{\phi}(\Psi(x_0,t_0)),$$ where $$K(x_0,t_0;x,t)$$ is defined as before, and for each path $$\phi:[t_0,t]\rightarrow X$$ with the boundary condition $$\phi(t_0)=x_0 \ \phi(t)=x$$, the map $$\Pi_{\phi}:\wedge^{r}T^*_{x_0}X\rightarrow \wedge^{r}T^*_x X$$ is defined to be the parallel transport along $$\phi$$ determined by the connection $$\nabla$$. This sounds reasonable, but I am not sure about it. Also I don't know which connection $$\nabla$$ to use. Could anyone let me know the right form of propagation?

The configuration space for this theory has even coordinates $$x$$ and odd coordinates $$\psi$$ (it is $$\Pi TX$$ as a supermanifold). The Hilbert space is (the $$L^2$$ completion of ) the space of function in those variables, ie $$\Omega^\bullet (X)$$.
The evolution operator is a unitary operator $$U(t_0,t):\Omega^\bullet(X) \to \Omega^\bullet(X)$$. If we want to write it as an integral kernel, it will involve an integral over the whole configuration space (over $$\Pi TX$$, rather than just $$X$$). More explicitly, with $$\Psi(t_0) \in C^\infty(\Pi TX)$$, we have :
\begin{align} \Psi(t,x,\psi) &= \Big[U(t,t_0)\Psi(t_0)\Big](x,\psi) \\ &= \int_{\Pi TX}\text dx_0\text d\psi_0K(x_0,\psi_0,t_0;x,\psi,t)\Psi(t_0,x_0,\psi_0) \end{align}
The kernel can be defined by a path integral as : $$K(x_0,\psi_0,t_0;x_1,\psi_1,t) = \int_{\phi(t_0) = x_0,\psi(t_0) = \psi_0}^{\phi(t) = x_1,\psi(t_1) = t_1}\mathcal D\phi\mathcal D\psi\mathcal D\bar \psi e^{-S_E(\phi,\psi,\bar\psi)}$$
We can also make sense of $$K(x_0,t_0;x,t)$$ as a linear map $$\bigwedge^\bullet T_{x_0}^* X \to \bigwedge^\bullet T_{x}^* X$$. Explicitely, for $$\alpha \in \bigwedge^\bullet T_{x_0}^* X = C^\infty(\Pi T_{x_0}X)$$, we have : $$(K(x_0,t_0;x,t) \alpha)(\psi) = \int_{\Pi T_{x_0}X}\text d\psi_0 K(x_0,\psi_0,t_0;x,\psi,t)\alpha(\psi_0)$$