TITLE: Splitting of the double tangent bundle into vertical and horizontal parts, and defining partial derivatives
QUESTION [9 upvotes]: 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.

REPLY [4 votes]: Denote $E=TM$. We have a vertical projection $P_{(x,v)}:T_{(x,v)}E\to E_x$ determined by the splitting. The covariant derivative of a vector field $(\gamma,u)$, regarded as a curve in $E$, is the vertical projection of its derivative. We also have linear maps $d\pi:T_{(x,v)}E\to T_xM$ (where $\pi:E\to M$ is the bundle projection), the horizontal lift $h:T_xM\to H_{(x,v)}$ such that $h\circ d\pi=id_H$, and the natural isomorphism $i:T_xM=E_x\to V_{(x,v)}$ (which is the only thing depending on the fact that $E$ is the tangent bundle and not an arbitrary vector bundle). In this notation, we have $\xi=h\circ d\pi(\xi)+i\circ P_{(x,v)}(\xi)$ for every $\xi\in T_{(x,v)}E$.
Define $\nabla_x A=P_{A(x,v)}\circ (dA)\circ h$ and $\nabla_v A=P_{A(x,v)}\circ  (dA)\circ i$ where $dA:T_{(x,v)}E\to T_{A(x,v)}E$ is the ordinary derivative. Then, for a vector $\xi\in T_{(x,v)}E$ (the derivative of our vector field $(\gamma,u)$) we have
$$
\begin{aligned}
 \nabla_t (A(\gamma,u)) &= P_{A(x,u)}\circ dA(\xi) =  P_{A(x,u)}\circ dA \circ h\circ d\pi(\xi) +P_{A(x,u)}\circ dA \circ i\circ P_{(x,v)}(\xi) \cr
 &=(\nabla_xA)(d\pi(\xi)) + (\nabla_vA)(P_{(x,v)}(\xi)) =(\nabla_xA)\cdot\dot\gamma +  (\nabla_vA)\cdot\nabla_t(\gamma,u) .
\end{aligned}
$$