I have difficulty understanding the following argument because it seems that many details are swept under the rug and I am looking for a detailed rigorous exposition on this (I'll try to make clear where I am especially looking for details).
Let $G$ be a Lie group, and $H$ be a closed subgroup, which is an embedded submanifold by closed subgroup theorem. Correct me if I mess up the facts of this part: $H$ is an integral manifold of an involutive distribution $D$, the collection of cosets $\{gH\}$ forms a foliation, and for the distirbution $D$ there are defining global left-invariant 1-forms $\omega^1,...,\omega^m$ (i.e. $T_g(gH)=\cap \ker (\omega^i_g)$ for every $g\in G$). (I don't need more details on this paragraph if I these statements are right, since I have the explanation somewhere).
Now, my book claims that $\omega^1,...,\omega^m$ are also (candidates for) 1-forms on $G/H$.
I am uncomfortable with the last statement, and here is where I need the details. It technically cannot be that $\omega^i$ is a form on $G/H$ - it has to be some 1-form that is induced "naturally" from $\omega^i$. What is it? What is a 1-form on $G/H$ or the tangent space of a point in $G/H$ anyway (described starting from 1-form on $G$ and the tangent spaces of points in $G$ and $H$)? I foresee some form of "pick a representative" argument (this is my guess), but I am not able to flesh out any details.
Let $\overline{g}\in G/H$ be represented by $g\in G$, and let $\overline{v}\in T_{\overline{g}}G/H$. Choose $v\in T_gG$ such that $d\pi_g(v)=\overline{v}$, and define$$\overline{\omega}_{\overline{g}}^i(\overline{v})=\omega_g^i(v).$$Independence of the tangent vector $v\in T_gG$ follows from $T_ggH\subset\ker(\omega_g^i)$. Independence of the representative $g$ follows from left invariance of $\omega^i$.