Using $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b}(\mathbf{a} \cdot \mathbf{c}) - \mathbf{c}(\mathbf{a} \cdot \mathbf{b})$ which is easy to prove we have
$$ \mathbf{k} \times (\mathbf{k} \times \mathbf{\mathbf{G}}) = \mathbf{k}(\mathbf{k} \cdot \mathbf{G}) - \mathbf{G} (\mathbf{k} \cdot \mathbf{k}) $$
so that
\begin{align}
\mathbf{F}(\mathbf{r}) &= \int \mathbf{G}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{r}} d^3 \mathbf{k} \\
&= \int \mathbf{G}(\mathbf{k}) \frac{\mathbf{k} \cdot \mathbf{k}}{||\mathbf{k}||^2} e^{i\mathbf{k} \cdot \mathbf{r}} d^3 \mathbf{k} \\