The bundle $G_2 \to Spin(7) \to S^7$ is nontrivial, indeed. The formula above $$g_{S^7} = \exp\left\{ \frac 12 \xi_j \phi_{jkl} \Gamma_k \Gamma_l \right \} \qquad \qquad (1)$$ ($\Gamma_j$ are the matrices $8 \times 8$ satisfying the Clifford algebra) is a local section, but not a global section of this bundle. It can be proven in quite explicit "physical" way. I do so in 2102.07415 [hep-th]. The digest of the proof is the following.
(1) is a section because
(i) Seven generators $T_j = \phi_{jkl} \Gamma_k \Gamma_l$ are orthogonal to 14 generators of $G_2$ and
(ii) (1) parameterize the whole $S^7$: the point $\xi_j = 0$ is its north pole with $g= \mathbb{1}$ and, if $\|\xi_j\| = \pi$, one obtains $g = - \mathbb{1}$, and this is the south pole.
However, (1) is not a global section by the following not quite trivial reason:
A subgroup $G_2 \subset Spin(7)$ are the matrices that do not transform a particular 8-component real spinor $\psi_0$. Consider the action of the matrices (1) on this spinor. One can show that $g \psi_0 = \cos(3\alpha) \psi_0 + \sin(3\alpha) \psi_1$, where $\alpha = \|\xi_j\|$ and $\psi_1$ is some other spinor. If $\alpha = 0, \pm 2\pi/3$, $\psi_0$ is left intact. Thus, many elements of (1) belong to one and the same fiber of the bundle and hence (1) cannot be a section.