Let $M$ be a manifold and $g$ a metric on $M$.
Let $TM$ denote the tangent bundle of $M$, and denote points in $TM$ by $(x,v)$ where $v \in T_xM$.
The Levi-Civita connection of $(M,g)$ induces a splitting of the double tangent bundle $TTM = V \oplus H$, where $V$ is the vertical distribution, defined by $V_{(x,v)} = T_{(x,v)}T_xM$ (i.e. the tangent space to the fibre), and $H_{(x,v)}$ is the horizontal distribution, which is determined by the connection.
Suppose $A:TM \rightarrow TM$ is a map such that $A(x,v)\in T_xM$ (so the map $A(x,\cdot)$ is a map from $T_xM$ to itself for all $x \in M$).
How does one use the splitting described above to define "partial derivatives" $\nabla_xA$ and $\nabla_vA$, which should be maps:
$(\nabla_xA)(x,v):T_xM \rightarrow T_xM$,
$(\nabla_vA)(x,v):T_xM \rightarrow T_xM$.
These should have the property that if $\gamma(t)$ is a curve on $M$ and $u(t)$ is a vector field along $\gamma$ (so $u(t) \in T_{\gamma(t)}M$ for all $t$), and $\nabla_t$ denotes the covariant derivative along $\gamma$, then
$\nabla_t(A(\gamma,u)) = (\nabla_xA)(\gamma,u) \cdot \dot{\gamma} + (\nabla_vA)(\gamma,u) \cdot \nabla_t u$
(here on the LHS, $A(\gamma,u)$ is itself a vector field along $\gamma$, so the notation $\nabla_t(A(\gamma,u))$ is meaningful).
The expression above "makes sense" intuitively, but I can't get the formalism to work properly.
Best Answer
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.