Suppose $\pi: E \rightarrow M$ is a vector bundle over a manifold $M$. Suppose that
$$k: E \times_M E \rightarrow \mathbb{R}$$
is a bundle metric on $E$ and that
$$g: TM \times_M TM \rightarrow \mathbb{R}$$
is a Riemannian metric on $M$.
Then if I have a connection $\nabla: \Gamma(TM) \times \Gamma(E) \rightarrow \Gamma(E)$ on $E$, I can write a splitting:
$$TE = HE \oplus VE \cong TM \oplus E$$
My question is, can I use these structures to define a Riemannian metric on the total space $E$? Also, are any of these three things unnecessary?
Here is my guess about how to construct the metric on the total space $E$:
$$G: TE \times_E TE \rightarrow \mathbb{R}$$
via
$$G_e(u,w) = k_p(u_v,w_v) + g_p(u_h , w_h)$$
where $u_v$ is the vertical projection of $u$ and $u_h$ is the horizontal projection of $u$ (so like $d\pi(u)$) and $p = \pi(e)$
If $\pi : E \to M$ is a smooth vector bundle, then there is a short exact sequence of smooth vector bundles on $E$
$$0 \to VE \to TE \to HE \to 0 \tag{$\ast$}$$
where $VE := \ker d\pi$ and $HE := TE/VE$. In particular, $VE$ is a subbundle of $TE$, whereas $HE$ is a quotient bundle of $TE$. Note that $d\pi : TE \to TM$ factors through $HE$, and as $\pi$ is a submersion, the induced map $HE \to TM$ is an isomorphism covering $\pi$, so $HE \cong \pi^*TM$. On the other hand, the vector bundle structure gives rise to a canonical isomorphism $VE \cong \pi^*E$; see this question.
A connection $\nabla$ on $E$ induces an injective map $\pi^*TM \to TE$ via parallel transport. Denote the image of the composition $HE \to \pi^*TM \to TE$ by $(HE)_{\nabla}$. Note that $(HE)_{\nabla} \cong HE$, but $(HE)_{\nabla}$ is a subbundle of $TE$, whereas $HE$ is a quotient bundle of $TE$. The composition $HE \to \pi^*TM \to TE$ is a right splitting of $(\ast)$ with image $(HE)_{\nabla}$, so we have
$$TE = VE\oplus(HE)_{\nabla} \cong VE\oplus HE \cong \pi^*E\oplus\pi^*TM \cong \pi^*(E\oplus TM).$$
Note that a Riemannian metric on $M$ is precisely a bundle metric on $TM$, so I will only refer to bundle metrics. If $E$ and $TM$ are equipped with bundle metrics, then so is $\pi^*(E\oplus TM)$. We can then pullback the bundle metric on $\pi^*(E\oplus TM)$ by the isomorphism $\Phi : TE \to \pi^*(E\oplus TM)$ to obtain one on $TE$.
Did we really need the connection? The connection gave us a particular splitting $TE = VE\oplus(HE)_{\nabla}$, but regardless of the connection, we have $TE \cong \pi^*(E\oplus TM)$ and hence we can construct a bundle metric on $TE$ from bundle metrics on $E$ and $TM$. However, the isomorphism $\Phi : TE \to \pi^*(E\oplus TM)$ depends on $\nabla$, so the particular bundle metric you obtain on $TE$ will depend on the choice of connection.