I'm in the final stages of proving the desired proposition, but I'm having difficulties showing that some vector fields are smooth. I'm given next to nothing about the division algebra on $\mathbb{R}^n$, which is where much of the difficulty for me lies. All we have is that multiplication is $\mathbb{R}$-bilinear. We do not have that the algebra is commutative, associative, or has $1$.
My argument is as follows. Express $S^{n-1} = \{x \in \mathbb{R}^n : ||x|| = 1\}$. For all $x \in \mathbb{R}^n$, the map $f_x (a) = \frac{x \cdot a}{||x \cdot a||}$ is a diffeomorphism $S^{n-1} \to S^{n-1}$. Fix $e = (1,0,\dots,0) \in S^{n-1}$, and choose a basis $\{\xi_{i,e} : i = 1, \dots, n\}$ of $T_eS^{n-1}$. For each $y \in S^{n-1}$, let $y'$ be the unique solution to the equation $y' \cdot e = y$. Then $T_ef_{y'} = Tf_{y'}|_{TeS^{n-1}}$ induces a linear isomorphism $T_eS^{n-1} \to T_{f_{y'}(e)}S^{n-1} = T_yS^{n-1}$ and thus maps $\{\xi_{i,e}\}$ to a basis of $T_yS^{n-1}$. Define the vector field $\xi_i: M \to TM$ by $\xi_i(a) = \xi_{i,a} = T_ef_{a'}(\xi_{i,e})$. It's clear that the $\xi_i$'s satisfy the requirements for being an $n$-frame, and now all I need to do is show that each $\xi_i$ is smooth. Since $\xi_i$ is defined somewhat indirectly, and relies on solving the equation $a' \cdot e = a$, I don't know if the solutions vary smoothly as $a$ varies smoothly (in coordinates).