at the end I'm not sure if you want $G$ to be a standard gaussian or a Rademacher  ?

If $G$ were to be a Rademacher, then consider the function $f$ such that $f(-1)=f(1)=0$ and $f(x)=1$ if $x\not\in\{-1,1\}$. Then the right term is equal to $0$. However if $U$ is not too trivial then the left term is positive. (example : $n=2$ take $U$ a rotation matrix of angle $\pi/3$). The idea is that the coordinates of $U^TX$ will not be Rademacher variables (a priori), so then we can play on the support of the measures.

If $G$ were to be a standard gaussian. Then take $U=I_n$ and consider the function $f$ such that $f(1)=f(-1)=1$ and $f(x)=0$ for all $x\not\in \{1,-1\}$. The left term will be positive while the right term is zero. (I think you can adapt this phenomenon if $U$ is general  : take a similar $f$ (that will depend on $U$) with an appropriate discrete support which will be all possible realizations of the coordinates of $U^TX$).

I think you will have to impose some hypothesis on $f$. You already checked that it was ok for $f(x)=x^2$ and I believe you can adapt that proof to show that your intuition is true if you have a condition like : $\alpha_1 x^2 \leq f(x)\leq \alpha_2 x^2$ (for some constants $0<\alpha_1\leq\alpha_2$). It may be true for a more general class of functions (however the $\alpha(n)$ may depend on the $f$)

Hope this helped  !

At the end I'm not sure if you want $G$ to be a standard Gaussian or a Rademacher?

If $G$ were to be a Rademacher, then consider the function $f$ such that $f(-1)=f(1)=0$ and $f(x)=1$ if $x\not\in\{-1,1\}$. Then the right term is equal to $0$. However if $U$ is not too trivial then the left term is positive. (example : $n=2$ take $U$ a rotation matrix of angle $\pi/3$). The idea is that the coordinates of $U^TX$ will not be Rademacher variables (a priori), so then we can play on the support of the measures.

If $G$ were to be a standard Gaussian, then take $U=I_n$ and consider the function $f$ such that $f(1)=f(-1)=1$ and $f(x)=0$ for all $x\not\in \{1,-1\}$. The left term will be positive while the right term is zero. (I think you can adapt this phenomenon if $U$ is general: take a similar $f$ (that will depend on $U$) with an appropriate discrete support which will be all possible realizations of the coordinates of $U^TX$).

I think you will have to impose some hypothesis on $f$. You already checked that it was ok for $f(x)=x^2$ and I believe you can adapt that proof to show that your intuition is true if you have a condition like : $\alpha_1 x^2 \leq f(x)\leq \alpha_2 x^2$ (for some constants $0<\alpha_1\leq\alpha_2$). It may be true for a more general class of functions (however the $\alpha(n)$ may depend on the $f$).

Hope this helped!