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 := \dot{z}(0)$$\xi_w$ is the unique vector whose horizontal component is $w$ and whose vertical component is zero (i.e. $\xi^h = w$ and $\xi^v = 0$), then we define
$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi))^h$$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi_w))^v$,
$(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\xi))^v$ 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 is,
$d_{(x,v)}A(\xi) \simeq ((\nabla_xA)(x,v)(w),(\nabla_vA)(x,v)(y))$,
and these two maps have the properties you're looking for.