The following is a counterexample: $$\Delta_n(x,y)=\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}+e^{-x^2-(y-n)^2},$$ $$g_1(x)=g_2(x)=x^2,\quad f(x)=e^{-x^2/2}.$$

Indeed, the first summand in the above expression for $\Delta_n(x,y)$ is a standard weak* approximation of $\delta(x-y)$. The second summand in the expression for $\Delta_n(x,y)$ is a weak* approximation of $0$, by (say) dominated convergence.

As for your second displayed equality, one way to see that it fails to hold is to compute explicitly both integrals there, which is easy to do. In particular, the limit on the left-hand side of that equality is $\infty$.

The following is a counterexample: $$\Delta_n(x,y)=\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}\,e^{-(x^2+y^2)/n} +e^{-x^2-(y-n)^2},$$ $$g_1(x)=g_2(x)=x^2,\quad f(x)=e^{-x^2/2}.$$

Indeed, the term $\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}$ in the above expression for $\Delta_n(x,y)$ is a standard weak* approximation of $\delta(x-y)$, and $e^{-(x^2+y^2)/n}\to1$ uniformly over $(x,y)$ in any given bounded set. The second summand in the expression for $\Delta_n(x,y)$ is a weak* approximation of $0$, by (say) dominated convergence.

As for your second displayed equality, one way to see that it fails to hold is to compute explicitly both integrals there, which is easy to do. In particular, the limit on the left-hand side of that equality is $\infty$.

I think the following is a counterexample: $$\Delta_n(x,y)=\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}+e^{-x^2-(y-n)^2},$$ $$g_1(x)=g_2(x)=x^2,\quad f(x)=e^{-x^2/2}.$$

The following is a counterexample: $$\Delta_n(x,y)=\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}+e^{-x^2-(y-n)^2},$$ $$g_1(x)=g_2(x)=x^2,\quad f(x)=e^{-x^2/2}.$$

Indeed, the first summand in the above expression for $\Delta_n(x,y)$ is a standard weak* approximation of $\delta(x-y)$. The second summand in the expression for $\Delta_n(x,y)$ is a weak* approximation of $0$, by (say) dominated convergence.

As for your second displayed equality, one way to see that it fails to hold is to compute explicitly both integrals there, which is easy to do. In particular, the limit on the left-hand side of that equality is $\infty$.