Does $\omega \wedge \mathrm{d} \omega=0$ (where $\omega$ is a non-vanishing $1$-form) imply $\mathrm{d} \omega \in \langle \omega\rangle$? Let $\omega$ be a non-vanishing (for clarification: nowhere vanishing) smooth $1$-form on a smooth manifold $M$, if $\mathrm{d}\omega \wedge \omega =0$, do we already have $\mathrm{d}\omega= \sum a_i \wedge \omega$ for some $1$-forms $a_i$?
 A: No: $M=\mathbb R^2,\: \omega=xd y$   
Edit
In the first version of the question "non vanishing" was in brackets and I interpreted the question as "not identically zero but maybe having zeros", as is indeed the case in my example..
A: 
But, being a beginner in differential forms on smooth manifolds: can we extend to a basis globally?

In general, no. On parallelizable manifolds, you can extend to a global basis in the sense that $(\omega(p),\omega_2(p),\dotsc,\omega_k(p))$ are a basis of the cotangent space at every point, but not on other manifolds.

Or does this suffices locally?

If your definition of manifold includes paracompactness (or second countability, which implies paracompactness for locally Euclidean Hausdorff spaces), then there exist smooth partitions of unity subordinate to every open cover of $M$.
You can then extend $\omega$ to a local frame on every coordinate patch $U$
 - meaning you find smooth $1$-forms $\omega_2,\dotsc,\omega_n$ on $U$ such that $\omega(p),\omega_2(p),\dotsc, \omega_n(p)$ is a basis of $T_p^\ast M$ for all $p\in U$ - and the condition $d\omega \wedge \omega = 0$ then implies that, with smooth functions $\beta_k$, we have
$$d\omega = \underbrace{\sum_{k = 2}^n \beta_k\cdot \omega_k}_{\alpha_U}\wedge \omega$$
on $U$. Cover $M$ with a family $\mathfrak{U}$ of such coordinate patches and find a subordinate smooth partition of unity $\{ \varphi_U : U \in \mathfrak{U}\}$. Then you have
$$d\omega = \sum_U \varphi_U\cdot d\omega = \sum_U \varphi_U (\alpha_U\wedge \omega) = \Biggl(\sum_U \varphi_U\cdot \alpha_U\Biggr)\wedge \omega.$$
