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 or 6.11 in that book.

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 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.

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 or 6.11 in that book.

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 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.

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 or 6.11 in that book.

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 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.

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 or 6.11 in that book.

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 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.

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 $$ T\operatorname{Hom}(E^*,F) = \operatorname{Hom}(E^*,TF) \xrightarrow{\operatorname{Hom}(E^*,\pi_F)} \operatorname{Hom}(E^*,F). $$ Note the second vector bundle structure $$ \operatorname{Hom}(E^*,TF)\xrightarrow{\operatorname{Hom}(E^*,T(p_F))} \operatorname{Hom}(E^*,TM), $$ see 8.12 of this book or 6.11 in that book.

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 of this book 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 is not natural.

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 or 6.11 in that book.

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 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.