Show that G is not a PRF
If the adversary takes $x, y \in \{0, 1\}^n$ such that $(0 \parallel x) = (y \parallel 1)$ then

$$G_K(x) = F_K(0 \parallel x) \parallel F_K(x \parallel 1)$$
$$G_K(y) = F_K(0 \parallel y) \parallel F_K(y \parallel 1) = F_K(0 \parallel y) \parallel F_K(0 \parallel x)$$

The left half of $G_K(x)$ is equal to the right half of $G_K(y)$, which lets the adversary distinguish $G$ from a random function.