What can we say about a vector bundle with trivial unit sphere bundle?

The classifying space for $S^k$ bundles would be $\mathrm{BDiff}(S^k)$. See this overview article by Hatcher for what's known about these:

For $k \leq 3$ it's known that $\mathrm{Diff}(S^k) \simeq O(k+1)$ under the inclusion map, with the case $k = 3$ due to Hatcher, so a trivial $S^k$ bundle implies the original vector bundle is trivial for $k \leq 3$.

The homotopy type of $\mathrm{Diff}(S^4)$ is not known.

Update:

However, as stated in that article, we do have $\mathrm{Diff}(S^k) \simeq O(k+1) \times \mathrm{Diff}(D^k\ rel\ \partial)$ (again under the map?), which implies $B\mathrm{Diff}(S^k) \simeq BO(k+1) \times B\mathrm{Diff}(D^k\ rel\ \partial)$, which implies that if a sphere bundle coming from a vector bundle is trivial, then the original vector bundle is trivial (all assuming the ? above).

---

Working on $\mathrm{Diff}(S^k) \simeq O(k+1) \times \mathrm{Diff}(D^k\ rel\ \partial)$:

Hatcher, in his proof that $\mathrm{Diff}(S^3) \simeq O(4)$ states that this is "well-known and easily shown."

Consider the point $p = (0,0,\ldots,0,1) \in S^k \subset \mathbb{R}^{k+1}$ and let $\mathrm{incl}: S^k \hookrightarrow \mathbb{R}^{k+1}$ be the standard inclusion. One gets $f: \mathrm{Diff}(S^k) \rightarrow GL_{k+1}(\mathbb{R})$ given by $\phi \mapsto [d(\mathrm{incl}\circ\phi)_p |\, \mathrm{incl}\circ\phi(p)]$, where this last is a block matrix with $\mathrm{incl}\circ\phi(p)$ a column vector. That is, the map is "evaluate the derivative at a point and adjoin the normal vector" (also discussed by Hatcher in the above paper).

Next we have $GL_{k+1}(\mathbb{R}) \simeq O(k+1)$ by Gram-Schmidt.

We get $f \circ \iota = \mathrm{id}_{O(k+1)}$ for $\iota: O(k+1) \hookrightarrow \mathrm{Diff}(S^k)$ the inclusion, so, at least so far we do have the "compatibility with the map from $O(k+1)$" that we needed above.

The map $f$ is a fibration with fiber "diffeomorphisms of $S^k$ which fix $p$ and have $d\phi_p = I$" which is homotopy equivalent to $\mathrm{Diff}(D^k\ rel\ \partial)$.

I'm not quite sure how we get it homotopy equivalent to the product.