vertical component of Lie bracket Let $f:X\to Y$ be a submersion equipped with a connection given by a vertical projection $\mathrm V$. Let $\vec v_1,\vec v_2$ be vector fields on $Y$ with unique horizontal lifts $\vec u_1,\vec u_2$.
I have read the vertical component $\mathrm V([\vec u_1,\vec u_2](x))$ of the Lie bracket at a point $x\in X$ depends only on $\vec u_1(x),\vec u_2(x)$, i.e $\vec v_1(fx),\vec v_2(fx)$.
Question. What is the intuition behind this fact and how can I prove it?
 A: First the calculation:
Let $z_1,...,z_n$ be a frame of $TY$ in some neighbourhood of a point and let $z_1',...,z_n'$ denote their horizontal lifts. Now any vector-field $v$ in that neighbhourhood admits a unique expansion $v= \sum_i \alpha_i\,z_i$, here $\alpha_i$ are functions whose value at a point depend only on the value of $v$ at that point. What happens is that when you lift $v$ horizontally is that you just get $\sum_i \alpha_i'\, z_i'$, where $\alpha_i'$ are now functions $X\to\Bbb R$ given by $\alpha_i'(x) = \alpha_i (f(x))$. Now if you take the commutator of two horizontal lifts $v'=\sum_i\alpha_i' z_i'$ and $u'=\sum_i\beta_i' z_i'$ you get:
$$\sum_{ij}[\alpha_i' z_i', \beta_j' z_j' ]=\sum_{ij} \alpha_i'\beta_j' [z_i',z_j'] + \sum_{ij}(\beta_j' z_j'(\alpha_i')\,z_j' - \alpha_i' z_i'(\beta_j')\,z_j)$$
Now the second summand is horizontal and as such the vertical component is only contained in the first summand. But at an arbitrary point $x$ that summand only depends on the values of $\alpha_i(f(x)), \beta_j(f(x))$, which are determined by $u_{f(x)}$ and $v_{f(x)}$ and the value of $[z_i',z_j']_x$, which is independent of $u,v$. In other words it does not depend on how the fields $u, v$ look like in a neighbourhood of the point.
As to the intuition:
The way a horizontal field varies in fibre direction is uniquely determined by the value of the field at the base-point. Since the commutator describes the way two vector fields mutually change in a direction, the vertical component of the commutator describes the mutual vertical variation of the two fields. But the vertical behaviour is completely determined by the value at the base-point.
A: S.harp's answer is very instructive. Below is the proof I was looking for.
Let $X\overset{f}{\to} Y$ be a submersion equipped with an Ehresmann connection specified by a vertical projection $K$. Let $\vec u_1,\vec u_2$ be vector fields on $X$.
Proposition. The restriction of the composite $K\circ \mathcal L_{\vec u_1}$ to the $C^\infty _X$-module of horizontal vector fields is $C^\infty _X$ linear. Consequently, its action on $\vec u_2$ (which usually depends on the 1-jet of $\vec u_2$) depends only on the value of $\vec u_2$ (e.g by the equivalence of projective $C^\infty _X$-modules and $C^\infty $ vector bundles over $X$).
Proof. Use the product/Leibniz rule. $$\mathcal L_{\vec u_1}(f\cdot \vec u_2)=\mathcal L _{\vec u_1}f\cdot \vec u_1 + f\cdot \mathcal L_{\vec u_1}\vec u_2 $$
If $\vec u_2$ is horizontal then applying $K$ kills the first summand on the right, giving the desired result.
Corollary. If $\vec u_1,\vec u_2$ are both horizontal then the value $K[\vec u_1,\vec u_2]x$ is determined by $\vec u_1x,\vec u_2x$.
Proof. Apply the proposition twice, using also the fact Lie bracket is given by Lie derivative.
