This question was also asked [here](https://math.stackexchange.com/questions/4403054/tangent-bundle-of-a-tensor-product-bundle) on math-stackexchange.


Let $E\to M$ and $F\to M$ be vector bundles. The structure of their tangents $TE$ and $TF$ is well known. In particular, connectors map $K\_E: TE \to E\times\_M E$ and $K\_F: TF \to F\times\_M F$ induce isomorphisms $TE \simeq E \times\_M (E\oplus TM)$ and $TF \simeq F \times\_M (F\oplus TM)$ (I am using here a fibered product notation, rather than the equivalent pullbacks).


Consider now the vector bundle $E\otimes F\to M$. Its tangent can be characterized in the same way as the tangent bundle of every vector bundle. My question is whether there exists a canonical isomorphism of $T(E\otimes F)$ involving the tangent bundles $TE$ and $TF$. I would suspect a positive answer, which in particular looks like a "Leibniz rule". Also, what is the connector $K\_{E\otimes F}$ induced by $K\_E$ and $K\_F$? After all, covariant derivative of tensor products satisfy a Leibniz rule.


Here is yet an alternative formulation. How does the choice of horizontal bundles for $TE$ and $TF$ determine a horizontal bundle for $T(E\otimes F)$?


The same question can be asked regarding the vector bundle $\operatorname{Hom}(E,F)$.


In fact, the question seems reducible to the case of $M$ being a point, so that $E$ and $F$ are just vector spaces. Then, $TE \simeq E\times E$ and $TF \simeq F\times F$. If $E$ has dimension $m$ and $F$ has dimension $k$, then the tensor product $TE\otimes TF$ has dimension $4mk$, which is twice the dimension of $T(E\otimes F)$. How to proceed from here?



Tangent bundle of a tensor product bundle
Over a point:
$$
T(E\otimes F) = (E\otimes F)\oplus (E\otimes F) = E\otimes (F\oplus F)
$$
which is naturally isomorphic to $(E\oplus E)\otimes F$ using the canonical flip $E\otimes F = F\otimes E$. Likewise
$$
T\operatorname{Hom}(E,F) = \operatorname{Hom}(E,F)\oplus \operatorname{Hom}(E,F)=\operatorname{Hom}(E,TF)
$$
The Leibniz rule mixes the two representations: Consider first curve
$\sum\_i e\_i(t)\otimes f\_j(t)$; its velocity at $t=0$ is then
$$\Big(\sum\_i e\_i(0)\otimes f\_j(0), \sum\_i e\_i'(0)\otimes f\_j(0) + \sum\_i e\_i(0)\otimes f\_j'(0)\big).$$
Counting entries you have $2mk$.


It is more clear to consider a curve in terms of bases
$$
\sum\_{i,j} c\_{ij}(t)\; e\_i\otimes f\_j = 
\sum\_j\Big(\sum\_{i} c\_{ij}(t)\; e\_i\Big)\otimes f\_j 
= \sum\_{i} e\_i\otimes \Big(\sum\_jc\_{ij}(t)\;f\_j \Big),
$$  

then its derivate via (footpoint, speed vector) is
$\Big(\sum\_{i,j} c\_{ij}(0)\; e\_i\otimes f\_j, \sum\_{i,j} c\_{ij}'(0)\; e\_i\otimes f\_j\Big)$. You see that you can move the function part from left to right which explains the isomorphism above.


For vector bundles it is similar: the $TM$-part should be there only once.


Added:
======


Now let $p\_E:E\to M$ and $p\_F:F\to M$ be vector bundles. Then
$$E\otimes F = \operatorname{Hom}(E^\*, F) = \operatorname{Hom}(F^\*,E)$$
where the last natural isomorphism is via transpose using $E^{\*\*}=E$.
Then
$$
T\operatorname{Hom}(E^\*,F) = \operatorname{Hom}(E^\*,TF) \xrightarrow{\operatorname{Hom}(E^\*,\pi\_F)} \operatorname{Hom}(E^\*,F),
$$
where the middle ${\operatorname{Hom}}$ abuses notation and uses unsaid conventions.
Note the second vector bundle structure
$$
\operatorname{Hom}(E^\*,TF)\xrightarrow{\operatorname{Hom}(E^\*,T(p\_F))} \operatorname{Hom}(E^\*,TM),
$$
see 8.12 ff of [this book](https://www.mat.univie.ac.at/%7Emichor/dgbook.pdf) or 6.11 in [that book](https://www.mat.univie.ac.at/%7Emichor/kmsbookh.pdf).


Your next question is essentially, how to write the induced connector $K\_{E\otimes F}: T(E\otimes F) \to (E\otimes F)\times\_M (E\otimes F)$ whose kernel would identify the pullback of $TM$ to $E\otimes F$ with the horizontal bundle. See 19.12 ff of [this book](https://www.mat.univie.ac.at/%7Emichor/dgbook.pdf) for background. Here we need a name for the canonical isomorphism $\rho:E\otimes TF = F\otimes TE$ (abuse of notation here). Then $K\_{E\otimes F} = Id\_E \otimes K\_F + \rho \circ Id\_F\otimes K\_E \circ \rho$.


Note that the horizontal bundle is not natural.


A remark to the formulation at end of your question: $TE$ is NOT a vector bundle over $M$, it has two vector bundle structures
$$
TM \xleftarrow{Tp} TE \xrightarrow{\pi\_E} E,
$$
and the chart changes over $M$ are quadratic (like for the Christoffel symbols).
So $TE\otimes TF$ does make sense only with a lot of abuse of notation and unsaid conventions.


