Smoothness of a G-invariant K-form

Let G be a Lie group and $G\curvearrowright M$ a smooth homogenous action. I want to understand why a G-invariant k-form $\eta$ on M is smooth. I tried to prove it by finding a smooth frame on M of invariant vector fields (a similar process when we prove that invariant k-forms on G are smooth). I fixed a point p $\in M$ and $v \in T_pM$ and defined the vector field $X_v$ by $X_v(q) = g_*v$ where $g \in G$ is such that $g.p = q$. But I can't prove that $X_v$ is well defined.

Maybe I need other way to prove that $\eta$ is smooth.

You're not going to be able to find a global $G$-invariant smooth frame on $M$, for the simple reason that most manifolds don't admit any global frames, let alone $G$-invariant ones. (Think of the standard action of $\text{SO(3)}$ on $S^2$.) You might be able to make this idea work by finding a smooth invariant frame in a neighborhood of each point. (The construction on page 526 of my Introduction to Smooth Manifolds might be useful for this purpose.)

But here's a simpler approach you might try. Choose a point $p\in M$ and define $\pi\colon G\to M$ by $\pi(g) = g\cdot p$. First show that $\pi^*\eta$ is a left-invariant form on $G$, and therefore smooth; and then show that locally $\eta = \sigma^*(\pi^*\eta)$, where $\sigma$ is a smooth local section of $\pi$. The techniques you need are all explained in Chapters 20 and 21 of ISM.