I elaborate my comment to show you that probably convolution theorem won't help you here and certainly: $\mathcal{F} \{ g(\xi)\frac{\partial^{2} U}{\partial\xi^{2}} \} \neq \mathcal{F} \{ g(\xi) \} \mathcal{F} \{ \frac{\partial^{2} U}{\partial \xi^{2}} \}$. First of all, the convolution theorem introduces the $*$ operator as:
$$h(\mathbf{x}) = f(\mathbf{x}) * g(\mathbf{x}) = \int_{\Omega} f(\mathbf{x}^{'}) g(\mathbf{x} - \mathbf{x}^{'}) d^{n}\mathbf{x}^{'}$$
Where $f$ and $g$ are functions: $f, g: \mathbb{R}^{n} \rightarrow \mathbb{R}$ and $\mathbf{x} \in \mathbb{R}^{n}$.
In terms of Fourier transforms, the convolution theorem says:
$$\mathcal{F}\{ f(\mathbf{x}) g (\mathbf{x}) \} (\mathbf{k}) = \mathcal{F} \{ f \} (\mathbf{k}) * \mathcal{F} \{ g \} (\mathbf{k}) = \int_{\Omega_{\mathbf{k}}} \mathcal{F}\{f\}(\mathbf{k}^{'}) \mathcal{F}\{g\}(\mathbf{k} - \mathbf{k}^{'}) d^{3} \mathbf{k}^{'} \neq \mathcal{F} \{ f \} (\mathbf{k}) \mathcal{F} \{ g \} (\mathbf{k})$$
So, in your case:
$$\mathcal{F} \{ g(\xi) \frac{\partial^{2} U(\xi)}{\partial \xi^{2}} \} = \mathcal{F} \{ g \} * \mathcal{F} \{ \frac{\partial^{2} U}{\partial \xi^{2}} \} = \int_{\Omega_{k}} \mathcal{F}\{g\}(k^{'}) \mathcal{F} \{ \frac{\partial^{2} U}{\partial \xi^{2}} \} (k-k^{'}) dk^{'} = \\ -\int_{\Omega_{k}} (k-k^{'})^{2}\mathcal{F}\{g\}(k^{'}) \mathcal{F} \{ U \} (k-k^{'}) dk^{'}$$
I would say it's impossible or at least it's really difficult to use the above integral when you can't simply factorize the unknown $\mathcal{F} \{ U \} (k-k^{'})$ out of the integral. So, in my opinion this way is probably a dead end.
Update: You transformed your initial heat equation to this one:
$$\frac{\partial U}{\partial t} = \frac{\partial \xi}{\partial x} \frac{\partial}{\partial \xi} \Bigg( \frac{\partial \xi}{\partial x} \frac{\partial U}{\partial \xi}\Bigg)$$
The main problem as I described above is that you thought the $*$ operator is a simple algebraic multiplication, which is not. So, if I take Fourier transform from your equation, I would have:
$$\frac{\partial \mathcal{F} \{ U \}}{\partial t} = \mathcal{F}\{ \frac{\partial \xi}{\partial x} \} * \mathcal{F}\{ \frac{\partial}{\partial \xi}\Bigg(\frac{\partial \xi}{\partial x} \frac{\partial U}{\partial \xi} \Bigg) \} = \mathcal{F}\{ \frac{\partial \xi}{\partial x} \} * \Bigg (ik \mathcal{F} \{ \frac{\partial \xi}{\partial x} \frac{\partial U}{\partial \xi} \} \Bigg ) = \mathcal{F}\{ \frac{\partial \xi}{\partial x} \} * \Bigg( ik \mathcal{F}\{ \frac{\partial \xi}{\partial x} \} * \mathcal{F} \{ \frac{\partial U}{\partial \xi} \} \Bigg) = -\mathcal{F}\{ \frac{\partial \xi}{\partial x} \} * \Bigg ( (k\mathcal{F}\{ \frac{\partial \xi}{\partial x} \}) * (k\mathcal{F} \{ U \}) \Bigg )$$
So the final equation is:
$$\frac{\partial \mathcal{F} \{ U \}}{\partial t} = -\mathcal{F}\{ \frac{\partial \xi}{\partial x} \} * \Bigg ( (k\mathcal{F}\{ \frac{\partial \xi}{\partial x} \}) * (k\mathcal{F} \{ U \}) \Bigg )$$
Or in simpler form by taking $\mathcal{F} \{ \frac{\partial \xi}{\partial x} \} = \tilde{\xi}$ and $\mathcal{F} \{ U \} = \tilde{U}$:
$$\frac{\partial \tilde{U}}{\partial t} = -\tilde{\xi} * ((k \tilde{\xi}) * (k \tilde{U}))$$
The above equation is extraordinarily difficult to solve cause it's a integro-differential equation in comparison to your initial heat equation.
Update 2: I was thinking if it is really possible to use above equation practically and I think still there are some hopes!:
$$-\tilde{\xi} * ((k \tilde{\xi}) * (k \tilde{U})) = - \int_{\Omega_{k}} \int_{\Omega_{k}} k^{'} (k - k^{'}-k^{''}) \tilde{\xi}(k^{'}) \tilde{\xi}(k^{''}) \tilde{U}(k-k^{'}-k^{''},t) dk^{'} dk^{''}$$
Define:
$$\mathcal{K}(k^{'},k^{''};k,t) = k^{'} (k - k^{'}-k^{''}) \tilde{\xi}(k^{'}) \tilde{\xi}(k^{''}) \tilde{U}(k-k^{'}-k^{''},t)$$
So:
$$\frac{\partial \tilde{U}(k,t)}{\partial t} = -\int_{\Omega_{k}} \int_{\Omega_{k}} \mathcal{K} (k^{'},k^{''};k,t) dk^{'} dk^{''}$$
By using really basic approximation of integral by summation as well as Forward-Euler method to integrate in time:
$$\frac{\tilde{U}_{k}^{t+\Delta t} - \tilde{U}_{k}^{t}}{\Delta t} = - \sum_{k^{'}} \sum_{k^{''}} \mathcal{K}_{k}^{t}(k^{'},k^{''}) \Delta k^{'} \Delta k^{''}$$
So, your update equation would be:
$$\tilde{U}_{k}^{t+\Delta t} = \tilde{U}_{k}^{t} - \Delta t \Bigg( \sum_{k^{'}} \sum_{k^{''}} \mathcal{K}_{k}^{t}(k^{'},k^{''}) \Delta k^{'} \Delta k^{''} \Bigg)$$
In each time-step the summation or integral of right hand side is known because you know your initial condition and you can calculate the integral or summation for your initial condition as well. I have no idea how stable or unstable would be this scheme, it worth to try at least one time.