The answer is yes and you need transversality in some form. Embed, by WhitneyHere we use Whitneys embedding theorem, but a weaker statement would suffice. Embed $M \subset \mathbb{R}^m$. Thus there is a monomorphism $TM \to M \times \mathbb{R}^m$ of vector bundles, dualizing to an epimorphism $M \times \mathbb{R}^m \to T^{\ast} M$. Let $a_1,\ldots ,a_m$ be the images of the basis vectors. You can write any $k$-form on $M$ as a linear combination with $C^{\infty} (M)$-coefficients of the forms $a_{i_1} \wedge \ldots a_{i_k}$. Done.
The answer is yes and you need transversality in some form. Embed, by Whitney, $M \subset \mathbb{R}^m$. Thus there is a monomorphism $TM \to M \times \mathbb{R}^m$ of vector bundles, dualizing to an epimorphism $M \times \mathbb{R}^m \to T^{\ast} M$. Let $a_1,\ldots ,a_m$ be the images of the basis vectors. You can write any $k$-form on $M$ as a linear combination with $C^{\infty} (M)$-coefficients of the forms $a_{i_1} \wedge \ldots a_{i_k}$. Done.