EASY METHOD:
Suppose that $(p,\omega_p)\in Q \equiv T^*(M)$, where $p \in M, \omega_p \in T_p^*(M)$, represents a point in the cotangent bundle $Q$ whose fibres are the cotangent linear spaces at each point in the base manifold. We have the canonical fibre bundle projection,
\begin{align*} \pi: \quad Q &\to M \\ (p,\omega_p) &\mapsto p \ \end{align*}
which induces the pullback of 1-forms,
\begin{align*} \pi^*: \quad \Omega (M) &\to \Omega (Q) \\ \omega &\mapsto \pi^*(\omega)\equiv \theta. \ \end{align*}
In local coordinates, if we write $ \omega = \sum_i \alpha_i \ \text{d}x^i $, then it is straightforward to compute the local representation of the pullback:
$ \theta = \pi^*(\sum_i \alpha_i \ \text{d}x^i) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)\ ``=" \sum_i \alpha_i \ \text{d}x^i$
Note the quotation marks I used to emphasize that one should be careful about abuse of notation and use them smartly for convenience knowing that it should not be taken too literally.
ALTERNATIVE METHOD:
For a vector field $X \in \mathfrak{X}(Q)$, one could locally decompose it into a component parallel to the fibre $T_p^*(M)$ and another parallel to the base $M$ for each $p \in M$ :
$ X = \sum_j (X_{\parallel}^j\circ\pi)\frac{\partial}{\partial x^j} + \sum_i X_{\perp}^i \frac{\partial}{\partial \alpha^i} $
where we have again considered $ \omega = \sum_i \alpha_i \ \text{d}x^i $. By this decomposition, we mean that
$ \pi_*(X)= \sum_j X_{\parallel}^j \ \frac{\partial}{\partial x^j}$
And therefore, we have that
\begin{align*} \theta_{(p,\omega_p)}(X) = \pi_{(p,\omega_p)}^*(\omega_p)(X) &= \omega_p (\pi_{*_{(p,\omega_p)}}(X)) \\ &= \sum_i \alpha_i(p)\ \text{d}x^i (\sum_j X_{\parallel}^j(p) \ \frac{\partial}{\partial x^j}) \\ &= \sum_i (\alpha_i\circ\pi)(p,\omega_p)(X_{\parallel}^i\circ\pi)(p,\omega_p)\ \end{align*}
Thus: $ \theta (X) = \sum_i (\alpha_i\circ\pi) (X_{\parallel}^i\circ\pi) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)(X) $
which is the same result that we convinced ourselves of before.