Tensors are multilinear maps on vectors. Differential forms are a type of tensor.

Consider the 1-form $df$ for some function $f$. How does it act on a vector $\vv$? That's easy: by giving the directional derivative, namely \begin{equation} df(\vv) = \grad f\cdot\vv \end{equation} This is just the master formula in a new guise. Recall that the master formula tells us how to relate $df$ and $\grad f$, namely \begin{equation} df = \grad f\cdot d\rr \end{equation} and that a similar construction maps any vector field $\FF$ to a 1-form \begin{equation} F = \FF\cdot d\rr \end{equation} It is thus natural to define \begin{equation} (\FF\cdot d\rr) (\vv) = \FF\cdot\vv \end{equation} In an orthonormal basis we have \begin{equation} g_{ij} \,\sigma^j = \ee_i\cdot d\rr \end{equation} so that \begin{equation} g_{ij} \,\sigma^j(\ee_k) = \ee_i\cdot\ee_k = g_{ik} \end{equation} which forces \begin{equation} \sigma^j(\ee_k) = \delta^j{}_k \end{equation} Setting $\sigma_p=g_{pj}\sigma^j$ as before, we also have \begin{equation} g(\sigma^i,\sigma_p) = g_{pj} \,g(\sigma^i,\sigma^j) = g_{pj} \,g^{ij} = \delta^i{}_p = \sigma^i(\ee_p) \end{equation}

The action of higher rank forms can be defined similarly. Recalling that \begin{equation} g(\alpha\wedge\beta,\sigma\wedge\tau) = g(\alpha,\sigma) \,g(\beta,\tau) - g(\alpha,\tau) \,g(\beta,\sigma) \end{equation} we set \begin{equation} (\alpha\wedge\beta) (\vv,\ww) = \alpha(\vv)\,\beta(\ww) - \alpha(\ww)\,\beta(\vv) \end{equation} so that in particular \begin{equation} (\sigma^i\wedge\sigma^j) (\ee_p,\ee_q) = \sigma^i(\ee_p)\,\sigma^j(\ee_q) - \sigma^i(\ee_q)\,\sigma^j(\ee_p) = \delta^i{}_p \,\delta^j{}_q - \delta^i{}_q \,\delta^j{}_p \end{equation}