This is too long for a comment.
As André and Ryan have hinted at in the comments, what you have is a principal fibre bundle $G_2 \to \operatorname{Spin}(7) \stackrel{\pi}{\to} S^7$. Fibre bundles are locally trivial, so of course around every point of the 7-sphere there is a neighbourhood $U$ so that $\pi^{-1}U \cong U \times G_2$. The diffeomorphism depends on a choice of section: a way to assign to every point in $U$ a unique element of $\operatorname{Spin}(7)$.
What I know how to describe in terms of octonions (i.e., Cayley numbers) is the above fibre bundle, and maybe this helps you.
Let $\mathbb{O}$ denote the Cayley numbers. They form an 8-dimensional vector space with basis $e_1,\dots,e_8$, where $e_1,\dots,e_7$ are imaginary units and $e_8 = 1$. Let $L_i$ denote left multiplication by the imaginary unit $e_i$, for $i=1,\dots, 7$. The $L_i$ are endomorphisms of $\mathbb{O}$ which obey the Clifford relations $$ L_i \circ L_j + L_j \circ L_i = \begin{cases} - \operatorname{id} & i=j \cr 0 & i\neq j \end{cases}$$ whence they define a linear representation of the Clifford algebra $C\ell(7)$ on $\mathbb{O}$. (You could also use right multiplication and this would give the other inequivalent Clifford module of $C\ell(7)$. Both Clifford modules turn out to be equivalent under the spin group.)
The spin group $\operatorname{Spin}(7)$ naturally lives inside $C\ell(7)$, whence we also have a linear representation of $\operatorname{Spin}(7)$ on $\mathbb{O}$. This is nothing but the spinor representation $\operatorname{Spin}(7) \to \operatorname{SO}(\mathbb{O})$. The orbit of $1 \in \mathbb{O}$ under $\operatorname{Spin}(7)$ is the sphere of unit octonions, which we can identify with $S^7$. The stabiliser of $1$ in $\operatorname{Spin}(7)$ is precisely a $G_2$ subgroup. This then gives the principal bundle $$G_2 \to \operatorname{Spin}(7) \stackrel{\pi}{\to} S^7$$ with $\pi(g) = g \cdot 1$.