Consider a Cartan geometry $\pi: \mathcal{G} \to M$ with Cartan connection $\omega$ modelled on the Klein geometry $(G, H)$.

The Cartan connection is supposed to formalize what it means to "roll without slipping" the homogeneous space $G/H$ on the manifold $M$. I am wondering if the following is one correct way to intuitively think about this.

Let $U \subseteq M$ be an open set such that we have a local section $\sigma: U \to \mathcal{G}$ of the Cartan bundle. Then we can pull down the Cartan connection to get a $\frak{g}$-valued 1-form $\sigma^* \omega$ on $U$.

Suppose we pick a point of contact $o \in G/H$ with a point $x \in M$. Then if we pick a tangent direction $v \in T_x M$ we can imagine rolling the homogeneous space without slipping along the infinitesimal curve corresponding to the tangent vector. The Cartan form $\sigma^* \omega$ assigns to this infinitesimal curve $v$ an infinitesimal transformation $X:= (\sigma^* \omega)(v) \in \frak{g}$. Since $G$ acts on $G/H$, Then I would assume that the infinitesimal transformation $X$ tells us how the point of contact $o \in G/H$ changes as we roll along the infinitesimal curve $v$.

Thus, if the point of contact is initially $o \in G/H$, then our new point of contact with $v(d) \in M$ after rolling for infinitesimal time $d$ along the infinitesimal curve $v$ will be $X(d).o \in G/H$, where the infinitesimal action of $\frak{g}$ on $G/H$ here is induced by the standard left action of $G$ on $G/H$.

I am wondering whether this reasoning is a correct way to think about rolling without slipping the homogeneous space along $M$. If not, is there some other way I can think about it geometrically?

Note that while this statement in terms of infinitesimal transformations is a bit fuzzy and "handwavy" in classical differential geometry where we do not have infinitesimal objects, it is entirely rigorous in synthetic differential geometry, and we can embed the category of smooth manifolds fully and faithfully into whatever smooth topos we use to model SDG.