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.