Ok, finally figured it out -- the confusion basically arises from slightly nonstandard notation. First, however, let me review some important facts about differential forms (mostly following the discussion on Wikipedia). The application of a 2-form to a pair of vectors is often written as
$ \alpha \wedge \beta(u,v) = \alpha(u)\beta(v) - \alpha(v)\beta(u). $
Differential forms are often introduced in the scalar setting (e.g., $\alpha$ and $\beta$ are $\mathbb{R}$-valued) and so this expression makes sense: $\alpha(u)$ and $\beta(v)$ are both real-valued, so we can simply multiply them to get another real number.
More generally, however, suppose that $\alpha$ and $\beta$ take values in some vector space $E$. Then there may no longer be a natural or obvious way to "multiply" vectors, and so in general the best we can do is write something like
$ \alpha \wedge \beta(u,v) = \alpha(u) \otimes \beta(v) - \alpha(v) \otimes \beta(u), $
where $\otimes$ denotes the outer or tensor product (which is defined for any vector space $E$). Note, then, that the wedge product of two $E$-valued forms is not $E$-valued, but in fact $E \otimes E$ valued!
Logically, however, an element of $E \otimes E$ can be thought of as a pair of elements of $E$. So if we happen to have some binary product $\cdot: E \times E \rightarrow E$, then we can adopt the convention that any $(E \otimes E)$-valued 2-form "automatically" gets mapped to the $E$-valued 2-form
$ \alpha \wedge \beta(u,v) = \alpha(u) \cdot \beta(v) - \alpha(v) \cdot \beta(u). $
In the case of the vector area, the most (only?) natural binary product on $\mathbb{R}^3$ is given by the cross product, since it takes a pair of 3-vectors to a 3-vector. However (and here's the confusion), standard notation in this case would be
$\alpha \wedge \beta(u,v) = \alpha(u) \times \beta(v) - \alpha(v) \times \beta(u),$
and not
$\alpha \times \beta(u,v) = \alpha(u) \times \beta(v) - \alpha(v) \times \beta(u).$
(Personally I prefer to stick with the $\wedge$ notation because it is the only thing that gives the reader a clue that antisymmetrization occurs.)
So, with all that pedantic garbage out of the way, we can write a really pedantic proof of the statement from the question:
Consider the algebra bundle $E = (\mathbb{R}^3, \times)$ over a disk-like region $M \subset \mathbb{R}^2$ and let $f$ be an $E$-valued 0-form representing a smooth immersion; let $N$ be the unit normal field on $M$ induced by $f$ (also viewed as an $E$-valued 0-form). Finally, let $dA$ be the standard volume form on $M$. Noting that
$df \wedge df(u,v) = df(u) \times df(v) - df(v) \times df(u) = 2 df(u) \times df(v) = 2 N dA(u,v)$
for all pairs of vectors $u,v \in \mathbb{R}^2$ (since the cross product of two tangents points in the normal direction!), we have
$ \int_M N dA = \frac{1}{2} \int_M df \wedge df = \frac{1}{2} \int_M d(f \wedge df) = \frac{1}{2} \int_{\partial M} f \wedge df,$
as desired (up to a change in notation).