Let us write the isomorphism
$T_{(x,v)}TM = H_{(x,v)}TM \oplus V_{(x,v)})TM \cong T_xM \oplus T_xM$
by
$\xi \simeq (\xi^h,\xi^v)$,
so that $\xi^h \in T_xM$ and $\xi^v \in T_xM$.
Here the identification $H_{(x,v)}TM \cong T_xM$ is given by the restriction of $d_{(x,v)}\pi$ to $H_{(x,v)}TM$ (where $\pi:TM \rightarrow M$ is the projection), and the isomorphism $V_{(x,v)}TM \cong T_xM$ is canonical.
The key point is that under these identifications, if $z$ is a curve on $TM$, say $z(t)=(\gamma(t),u(t))$ then
$\dot{z}(0) \simeq (\dot{\gamma}(0),(\nabla_tu)(0))$.
So suppose $x \in M$ and $v,w,y \in T_xM$. Let $\gamma$ be a curve in $M$ such that $\gamma(0)=x$ and $\dot{\gamma}(0)=w$, and let $u$ be a vector field along $\gamma$ such that $u(0)=v$ and $(\nabla_tu)(0)=y$. Let $z(t)=(\gamma(t),u(t))$.
Think of $A$ as a map $TM \rightarrow TM$, so that the differential $dA$ is a map
$d_{(x,v)}A:T_{(x,v)}TM \rightarrow T_{(x,v)}TM$.
Then given $w\in T_xM$, if $\xi_w$ is the unique vector whose horizontal component is $w$ and whose vertical component is zero (i.e. $\xi^h = w$ and $\xi^v = 0$), then we define
$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi_w))^v$,
and similarly if $\zeta_w$ is the unique vector whose horizontal component is zero and whose vertical component is $w$ (i.e. $\zeta^h = 0$ and $\zeta^v = w$), then we define
$(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\zeta_w))^v$.
Then it follows that
$d_{(x,v)}A(\xi) \simeq ((\nabla_xA)(x,v)(w),(\nabla_vA)(x,v)(y))$,
and these two maps have the properties you're looking for.
There is nothing special about the tangent bundle: for any vector bundle $E \rightarrow X$ with projectivisation $\pi : \mathbb{P}E \rightarrow X$, and tautological bundle $\mathcal{L} \rightarrow \mathbb{P}E$, we have $T\mathbb{P}E \cong (\pi^*E / \mathcal{L}) \otimes \mathcal{L}^*$. In fact the base $X$ is not really involved, so you may as well work with the case where $X$ is a point, and $E$ is just a vector space $V$. Then the result follows from the standard Euler sequence
$$ 0 \rightarrow \mathcal{O}_{\mathbb{P}V} \rightarrow V \otimes \mathcal{O}_{\mathbb{P}V}(1) \rightarrow T\mathbb{P}V \rightarrow 0, $$
where we regard $V$ as a constant sheaf on $\mathbb{P}V$.
Added in response to comment: Explicitly, given a point $l \in \mathbb{P}V$, viewed as a line in $V$, and a vector $v \in T_l \mathbb{P}V$, we can lift $v$ (under the projection $V \setminus \{0\}\rightarrow \mathbb{P}V$) to a vector in $V/l$ at each point of $l \setminus \{0\}$. Moreover, this lift is equivariant under the action of $k^\times$ on $l\setminus \{0\}$, where $k$ is our ground field (e.g. $\mathbb{C}$). In other words, it defines a linear $V/l$-valued function on $l$.
Linear functions on $l$ are precisely elements of $\mathcal{L}^*_l$, and $V/l$ is precisely $(\pi^*V/\mathcal{L})_l$, so $v$ defines an element of $((\pi^*V/\mathcal{L}) \otimes \mathcal{L}^*)_l$.
Added in response to further comments: Even more explicitly, if $e_0, e_1, \dots, e_n$ is a basis for $V$, with corresponding coordinates $w_0, w_1, \dots, w_n$, then the $w_i$ form homogeneous coordinates on $\mathbb{P}V$. On the affine patch $w_0 \neq 0$, we have local coordinates given by $z_j = w_j/w_0$, for $j=1, \dots, n$. Given a point $l = [1: l_1: \dots: l_n] \in \mathbb{P}V$, the vector
$$ \sum_{j=1}^n v_j \frac{\partial}{\partial z_j} $$
is tangent to the curve $z_j(t) = l_j + tv_j$. This lifts to the curve in $V$ given by $w_0(t)=f(t)$, $w_j(t)=f(t)(l_j+tv_j)$ for $j\geq 1$, where $f$ is any smooth $\mathbb{C}^\times$-valued function. Its tangent at $t=0$, i.e. a lift of $v$ to $V$, is
$$ f'(0) \bigg( e_0 + \sum_{j=1}^n l_je_j \bigg) + \sum_{j=1}^n v_j e_j. $$
The ambiguity coming from $f'(0)$ is exactly what is killed when we quotient by $l$, so the lift to $V/l$ is represented by $\sum v_je_j$.
Explicit computation for $V=\mathbb{C}^2$: Take coordinates $w_0, w_1$ on $\mathbb{C}^2$, and let $U_j = \{w_j \neq 0\} \subset \mathbb{P}^1$. We have
$$(\pi^*V / \mathcal{L})_{[w_0 : w_1]} = \mathbb{C}^2 \Big/ \mathbb{C} \begin{pmatrix} w_0 \\ w_1 \end{pmatrix}.$$
On $U_0$ a complement to $\mathcal{L}$ in $\pi^*V$ is given by $\mathbb{C}e_1$; we can therefore take $[e_1]$ as a trivialising frame for the quotient bundle (where the square brackets denote the image in the quotient). Similarly on $U_1$ a frame is given by $[e_0]$. On $U_0 \cap U_1$ we have for all local holomorphic functions $f$ that
$$f e_1 = \frac{f}{w_1} \begin{pmatrix} w_0 \\ w_1 \end{pmatrix} - \frac{fw_0}{w_1} e_0 $$
so
$$ f[e_1] = -\frac{fw_0}{w_1}[e_0].$$
The transition function $\psi_{10}$ is therefore $-w_0/w_1$, which (up to the minus sign, which we could have eliminated by choosing $[-e_0]$ as our frame on $U_1$) is the transition function for $\mathcal{O}(1)$.
Best Answer
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.