$f^{*}$ is the complex conjugate and $\tilde{f}$ is the fourier transform of Ff$.
If $g(-\xi)=f(\xi)^{*}$ how does this imply $\tilde{g}(k)=\tilde{f}(k)^{*}$. This result just does not seem true by the property of the fourier transform that is if $f(ax)$ the fourier transform is $\frac{1}{a}\tilde{f}(\frac{k}{a})$.
If $h\colon\mathbb R\to\mathbb C$ is an integrable function, then $$\tag{1}\overline{\int_{\mathbb R}h(x)\mathrm dx}=\int_{\mathbb R}\overline{h(x)}\mathrm dx.$$
Using (1) with $h(x)=e^{-ikx}g(x)$, we obtain $$\overline{\widehat{g}(k)}=\int_{\mathbb R}e^{ikx}\overline{g(x)}\mathrm dx=\int_{\mathbb R}e^{ikx}f(-x)\mathrm dx.$$ We conclude using the substitution $t:=-x$ in the last integral.