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][1] or 6.11 in [that book][2].

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][1] 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
$$
TM \xleftarrow{Tp} TE \xrightarrow{\pi_E} E,
$$
so $TE\otimes TF$ does make sense only with a lot of abuse of notation and unsaid conventions.


  [1]: https://www.mat.univie.ac.at/~michor/dgbook.pdf
  [2]: https://www.mat.univie.ac.at/~michor/kmsbookh.pdf