Curves have $r$-th order contact iff their horizontal lifts have $r$-th order contact

Let $\pi: P \rightarrow M$ be a $G$-principal bundle. Let $\Phi \in \Omega^1(P, \mathfrak{g})$ be a principal connection on $P$. Let $I$ be an open interval with $0 \in I$. Consider two curves $\gamma_1$, $\gamma_2: I \rightarrow M$ with $\gamma_1(0)=\gamma_2(0)=x$. Fix $u \in P_x$. Denote by $\gamma_i^*:I \rightarrow P$ the uniquely determined horizontal lift of $\gamma_i$ with $\gamma_i(0)=u$ for $i=1,2$.

$\gamma_1$, $\gamma_2$ have contact of order $r$ in zero if and only if $\gamma_1^*$ and $\gamma_2^*$ have contact of order $r$ in zero.

Remember that two curves are said to have contact of order $r$ in a point, if there exist (equivalently: for all) local coordinates around that point, such that all derivatives up to the $r$-th derivative of the coordinate representations of the two curves coincide.

Note also, that $\gamma_1$, $\gamma_2$ have contact of order $r$ if and only if for any smooth function $\varphi: M \rightarrow \mathbb{R}$ we have $\varphi \circ \gamma_1 - \varphi \circ \gamma_2 = o(t^r)$.

The direction "$\Leftarrow$" is trivial, when using the second characterization.

I have problems showing the direction "$\Rightarrow$". I would like to use local coordinates $(x_1, \dots, x_n, y_1, \dots, y_m)$ around $u$, such that $\frac{\partial}{\partial x_i}$ are horizontal and $\frac{\partial}{\partial y_j} \in Ker (d \pi)$. Then the claim would follow, because $(x_1 \circ \pi ^{-1}, \dots, x_n \circ \pi^{-1})$ would be a chart of $M$ around $x$. (It needs some extra explanation, why $x_i \circ \pi^{-1}$ is well-defined, i.e. independent of the choice of the preimage under $\pi$) And we could check the first criterion for contact for the curves $\gamma_1^*$, $\gamma_2^*$: The last $m$ coordinates of all derivatives are zero for both curves. The first $n$ coordinates of all derivatives coincide with the derivative of the original curves $\gamma_1$, $\gamma_2$. (In the previously described coordinates respectively)

But why do such coordinates exist? I am not even sure they do exist...

Let $\sigma : M \supset U \rightarrow P$ be a local section. This defines a local trivialization of $P$, namely $\theta : (x, g) \mapsto \sigma (x) \cdot g$. Now write $\theta^{-1} \circ \gamma_i^* =(\gamma_i,g_i)$ for some $g_i : I \rightarrow G$. Then $0=\Phi ((\gamma_i^*)'(t))=\Phi ( d \theta (\gamma_i'(t),0))+\Phi(d\theta(0,g_i'(t)))$, i.e. $$\Phi ( d \theta (\gamma_i'(t),0))=-\Phi(d\theta(0,g_i'(t))). \tag{1}$$ We have $$\Phi ( d \theta (\gamma_i'(t),0))=\Phi(d \sigma (\gamma_i'(t)))\tag{2}$$ and $$\Phi(d\theta(0,g_i'(t)))=dL_{g_i^{-1}(t)} (g_i'(t)).\tag{3}$$ By assumption $\gamma_i'$ have contact of order $(r-1)$ at zero. Hence equation (2) gives that the left hand side of equation (1) has $(r-1)$-th order contact. This shows that also the right hand side of (3) has $(r-1)$-th order contact. This means however that the $g_i$ have contact of order $r$. (Check this by writing the curves $g_i$ in the map $\exp$)
So the $\gamma_i$ and $g_i$ have $r$-th order contact, hence $\theta (\gamma_i,g_i)$ also have $r$-th order contact.