p-forms as multilinear maps I'm studying differential geometry and am learning about differential forms. We have a very intuitive and simple way to understand 1-forms as linear maps on from the tangent space to the base field, defined as $$df(v) = vf$$ where $v$ is a vector field acting on $f$.
Now what my book does not mention and what I cannot seem to make up is how $p>1$ forms fit into this picture as multilinear forms. I know how to apply the exterior derivative and understand the exterior algebra, but it seems there is something missing. For example, for the case of a  2-form $df\wedge dg$, I would expect there to be a map that takes two vector fields and produces a real number. What is this map, both conceptually and syntactically? And how does this generalize to arbitrary p-forms?
 A: I'll write about the case of a single tangent space $V$, since this question is really about algebra.
In the quotient algebra construction, a form (an element of the exterior algebra) is a coset $[\omega] = \omega + I$, where $\omega$ is a $p$-tensor (multilinear map $V^p \to \mathbb F$) and $I$ is the two-sided ideal of the tensor algebra generated by all symmetric products $\xi \otimes \xi$, $\xi$ any tensor. The wedge product is just the multiplication induced by the tensor product; i.e. $[\alpha] \wedge [\beta] = [\alpha \otimes \beta]$.
Each of these cosets $[\omega]$ contains a unique totally antisymmetric tensor $\bar \omega$; i.e. a multilinear map $\bar\omega : V^p \to \mathbb F$ that changes by the sign of the permutation when you permute the inputs. Thus the exterior algebra (a quotient of the tensor algebra) can be naturally identified with the space of totally antisymmetric tensors (a subspace of the tensor algebra). This is very convenient for the purposes of differential geometry (it lets differential forms act on vectors), but the quotient definition is often preferred exactly because it gives you the wedge product for free.
Using this identification, we can take the wedge product and let it act on antisymmetric tensors: $\bar\omega \wedge \bar\alpha := \overline{\omega \wedge \alpha}$.
The wedge product of two one-forms then turns out to be $ \alpha \wedge \beta = \frac12 \left( \alpha \otimes \beta - \beta \otimes \alpha \right)$; i.e. $$(\alpha \wedge \beta)(u,v) = \frac12 \left( \alpha(u) \beta(v) - \beta(u) \alpha(v) \right).$$
Many authors define the wedge product on antisymmetric tensors without the factor of $\frac12$ - doing this makes the formulas simple, but makes the relationship back to the quotient algebra less natural.
In general the wedge product of $p$ $1$-forms $\theta_i$ turns out to be (as you'd expect) the total antisymmetrization of their tensor product:
$$ (\theta_1 \wedge \cdots \wedge \theta_p)(u_1, \ldots, u_p) = \frac1{p!}\det_{ij} (\theta_i(u_j)) = \frac1{p!} \sum_{\sigma} \theta_1 (u_{\sigma(1)}) \cdots \theta_p (u_{\sigma(p)})$$ where the sum is over all permutations of $\{ 1 , \ldots,  p \}$. (Those other authors I mentioned would omit the $1/p!$ term.)
If you want you can use this to derive the expression for the wedge product of two forms of arbitrary order - it gets even messier. (See e.g. Spivak's Calculus on Manifolds - one of the "other" authors.) Comparing this to our original definition of the wedge product, one can appreciate why the quotient construction is so nice to have.
A: In general, if a $p$-form $\omega:T_pM\times...\times T_pM\rightarrow\mathbb R$, can be written as exterior product of $p$ 1-form such as $\omega=\alpha_1\wedge...\wedge\alpha_p$. And if given vector fields $X_1,...,X_p$, we can explicitly compute 
$$\omega(X_1,...,X_p)=\det\left(\alpha_i(X_j)\right)$$
