Using the calculus of one forms prove this identity

Question

Given a map $\omega$ and a vector field $X$ on manifold $M$, prove that $$ d\omega(X)= X\omega $$ It is given that $d\omega$ is a one-form and $(d\omega(X))(p) = (d\omega)_p(X(p))$ for $p\in M$.

Here is my attempt

I don't have a definition of $X\omega$ but the book says that for a given parametrization, $f:U \to M$ on can write $$X =\sum_i a_i \frac{\partial}{\partial x_i} $$ such that $$X\omega =\sum_i a_i \frac{\partial\omega}{\partial x_i} $$

Then

$$(d\omega)_p(X(p))= \sum_i a_i \frac{\partial \omega}{\partial x_i}(p)=X\omega(p)$$

Questions:

1. is this correct?
2. what is the actual definition of $X\omega$