Suppose $B$ is a Riemannian manifold and $\pi: E \to B$ is a smooth vector bundle equipped with a metric.
Is there a natural Riemannian metric on $E$, i.e. a bundle metric on $TE\to E$?
It seems like there should at least be metrics on $TE$ which restrict to the original Riemannian metric on $TB \subset TE$. I tried to show that the metrics on $E \to B$ and $TB \to B$ give a splitting of the short exact sequence $VE \to TE \to \pi^* TB$ but ended up swimming in symbols.
As expected in the comment of @MikeMiller, I think the answer is "You need a connection on $E$", although I don't think that it has to be a linear connetion. As you pointed out in the question, for each point $u\in E$, there is an exact sequence $0\to V_uE\to T_uE\to T_{\pi(u)}B\to 0$. Since $E$ is a vector bundle $V_uE$ is canonically isomorphic to $E_u$, the fiber of $E$ over $u$. So the Riemannian metric on $B$ and the bundle metric on $E$ give you inner products on $T_{\pi(u)}B$ and $V_uE$, respectively.
Now an inner product on $T_uE$ which "fits into the sequence" is equivalent to specifying a subspace $H_uE$ in $T_uE$, which is complementary to $V_u$. (Given the inner product, take $H_uE$ to be $(V_uE)^\perp$, given the space $H_uE$, identify it with $T_{\pi(u)}B$ via the bundle projection, pull back the inner product and declare the sum to be orthogonal.) But a choice of complementary subspace is just a connection on the fiber bundle $\pi:E\to B$ (it is not neccesarily a linear connection on the vector bundle $E$).
In special situations, there may be a way to choose a specific connection by requiring some additional properties, but in general I don't see a way how to make a canonical choice.