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If $E \to B$ is a smooth vector bundle, then the tangent bundle $T E$ is a vector bundle over both $E$ and $T B$. (If you like, the latter structure is the derivative of the vector bundle structure on $E$.) If $F \to B$ is another smooth vector bundle and $\pi : T B \to B$ is the obvious projection, I have seen it written (e.g. here) that there is a canonical isomorphism $T(E \otimes F) \cong \pi^* E \otimes T F$ as vector bundles over $T B$.

Supposing that we can do this, we can in particular take $F = B \times \mathbb{R}$ (the trivial rank 1 bundle over $B$) and obtain a vector bundle isomorphism $$ \pi^* E \otimes_{TB} T(B \times \mathbb{R}) \to T E. $$ But since $T \mathbb{R} \cong \mathbb{R}^2$ canonically, there is a chain of canonical isomorphisms of vector bundles over $T B$ $$ \pi^* E \otimes_{TB} T(B \times \mathbb{R}) \cong \pi^* (E \otimes T \mathbb{R}) \cong \pi^* (E \oplus E). $$ (To avoid doubt, $E \otimes T \mathbb{R}$ just means the fiberwise tensor product of the vector bundle $E$ with the ordinary vector space $T \mathbb{R}$.)

In other words, we have $T E \cong \pi^*(E \oplus E)$ over $T B$. Of course this is locally (noncanonically) true, but globally such an isomorphism would e.g. define a canonical connection on any vector bundle $E$ by giving horizontal lifts of arbitrary vector fields on $TB$ to $TE$.

I imagine, then, that the precise statement is that connections on $E$ induce isomorphisms $T(E \otimes F) \cong \pi^* E \otimes T F$; intrinsically, how do we define this map? The answer I linked above also made reference to a canonical isomorphism $\rho : \pi^* E \otimes T F \to T E \otimes \pi^* F$, which appears to me to be in the same situation.

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I have explained the situation when there are connections on both $E$ and $F$ on MathOverflow over here. As I explain there, there is a surjective map $$ \tau : TE \otimes_{TB} TF \to T(E \otimes F) $$ which has a kernel (since the fiber dimension of the source is twice that of the target). As I also explain there, a connection on $E$ induces a direct sum decomposition of $TB$-bundles $TE \cong ZE \oplus_{TB} HE$ (with $ZE$ and $HE$ of the same dimension.

In our case, this means that $\tau$ splits as a sum of maps $$ \tau_Z : ZE \otimes_{TB} TF \to T(E \otimes F)\\ \tau_H : HE \otimes_{TB} TF \to T(E \otimes F). $$ The images $\tau_Z(ZE \otimes_{TB} TF)$ and $\tau_H(HE \otimes_{TB} ZF)$ are both equal to $Z(E \otimes F$), so the image of $\tau_Z$ is contained in the image of $\tau_H$. Since $\tau$ is surjective, by counting dimensions the only way this is possible is if $\tau_H$ is an isomorphism.

Of course, a connection on $E$ gives a corresponding horizontal lift isomorphism $\operatorname{hlift} : \pi^*E \to HE$, so (since $\tau_H \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF}) = \tau \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF})$) the composite $$ \mu := \tau \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF}) : \pi^*E \otimes TF \to T(E \otimes F) $$ is the canonical isomorphism we desire.


By the way, if $K_E : TE \to E$ and $K_F : TF \to F$ are both connector maps over here I show that the canonical connector $K_{E \otimes F}$ for $T(E \otimes F)$ is $$ K_{E \otimes F} \circ \tau = K_E \otimes \pi_F + \pi_E \otimes K_F. $$ Under the identification $\pi^*E \otimes TF \cong T(E \otimes F)$ above this means that the connector for $\pi^*E \otimes TF$ becomes \begin{align*} (X, e) \otimes \xi \in \pi^*E \otimes TF &\mapsto ( K_E \otimes \pi_F + \pi_E \otimes K_F) (\operatorname{hlift}(X,e) \otimes \xi)\\ &= K_E (\operatorname{hlift}(X,e)) \otimes \pi_F \xi + \pi_E (\operatorname{hlift}(X,e)) \otimes K_F \xi\\ &= 0 \otimes \pi_F \xi + e \otimes K_F \xi\\ &= (\pi_E \otimes K_F)((X, e) \otimes \xi). \end{align*}

In other words, the connector on $\pi^*E \otimes TF$ is just $\pi_E \otimes K_F$, with the connection on $E$ hidden in the identification $\mu : \pi^*E \otimes TF \to T(E \otimes F)$ itself.

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