What does the notation $g\cdot\omega$ mean in Spivak's Calculus on manifolds? In chapter $4$ (Integration on chains) of Spivak's Calculus on manifolds he says the following:

If $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is differentiable and $\omega$ be a $k$ form on $\mathbb{R}^m$  then
  $f^*(g\cdot\omega)=(g\circ f)\cdot f^*(\omega)$.

He said nothing about the $g$.
I think $g : \mathbb{R}^m \to \mathbb{R}$ differentiable. Then I guess we define $g\cdot\omega:\mathbb{R}^m\rightarrow \Lambda^k(\mathbb{R}^m)$
as 
$$(g\cdot\omega)(p)(v_1,\cdots,v_k)=g(v_1)g(v_2)\cdots g(v_k)\omega(p)(v_1,\cdots,v_k).$$
I am sure this does not make much sense but this is what I can think of as of now.
Please let me know more about this.

If $\omega$ is a differential $k$-form and $g$ is a smooth real-valued function then $g\cdot\omega$ (or just $g\omega$) is a differential $k$-form with
$$(g\cdot\omega)(p)(v_1, \dots, v_k) = g(p)\omega(p)(v_1, \dots, v_k).$$
What you wrote doesn't make sense as $p \in \mathbb{R}^m$ while $v_1, \dots, v_k \in T_p\mathbb{R}^m$, so $g(v_1), \dots, g(v_k)$ are not defined as $v_1, \dots, v_k$ are not elements of the domain of $g$.
More generally, for any smooth manifold $M$, given a differential $k$-form $\omega \in \Omega^k(M)$ and smooth function $g \in \mathcal{C}^{\infty}(M)$, $g\cdot\omega$ is again a differential $k$-form, defined as above. In fact, $\Omega^k(M)$ is a $\mathcal{C}^{\infty}(M)$-module.
