For a given function $f(x,y)$, its double integral is defined as:
$$\iint_R f(x,y)\;\mathrm{d}x\;\mathrm{d}y=\lim_{n\to\infty}\; \sum_{i=1}^{n }f(x_{i},y_{i})\Delta A_{i}$$
Where $\Delta A_{i}=\Delta x_{i}\Delta y_{i}$.
Later I was told it's $\mathrm{d}x\ \wedge \mathrm{d}y$ not $\mathrm{d}x\;\mathrm{d}y$. $\wedge$ being the wedge product.
This implies for instance that in polar, the differential with respect one integrates is given by:
$\mathrm{d}x\ \wedge \mathrm{d}y=r\mathrm{d}r\ \wedge \mathrm{d} \theta\ $
Whereas if it were the ordinary product $dxdy$ then we would have
\begin{align} dxdy & = \left(dr \cos \theta - r \sin \theta\ d\theta \right) \left( dr \sin \theta + r \cos \theta\ d\theta\right)\\ & = dr^2 \cos \theta \sin \theta - r^2 d\theta^2 \cos \theta\ \sin\ \theta + r\ dr\ d\theta\ (\cos^2 \theta\ - \sin^2\theta ) \end{align}
But nowhere in the above definition
$$\iint_R f(x,y)\;\mathrm{d}x\;\mathrm{d}y=\lim_{n\to\infty}\; \sum_{i=1}^{n }f(x_{i},y_{i})\Delta A_{i}$$
I see the wedge product. It Seems to me that $$\lim_{n\to\infty}\; \sum_{i=1}^{n }\Delta A_{i}=\iint \mathrm{d}A=\iint dxdy$$
Then where does the wedge product arise in the definition?
Short answer: it's the transformation formula.
Slightly longer answer, with some handwaving: The $n$-fold wedge product in $n$ dimensions is basically the same as the determinant (the determinant and the wedge product are both alternating n-Forms and that space is $1$ dimensional on $n$-dimensional vector spaces, so it's clear that the first is a multiple of the second).
Volume (area in two dimensions) transforms under linar transformations by multiplication with the determinant of the map associated with the transformation. If you look at a rectangle and apply a linear map to it the volume change is given by the determinant of the linear map. For differentiable transformation the determinant is replaced by the determinant of the differential of the transformation (up to sign, but that's a technical thing I'll ignore here). If you already came across the transformation formula you will have seen the Jacobian pop up:
$$ \int_\Omega f\circ \phi(x) \,|detD\phi|(x) \,dx= \int_{\phi(\Omega)} f(y) \, dy $$ here, $dx$ is actually an abbreviation for $dx^1 dx^2\cdots dx^n$, indicating the variables to integrate over. So the 'infinitesimal' area change is given by the determinant of the linearization of the transformation.
This is just the transformation behaviour of the wedge product, which is no surprise if you are aware of the remark I started with. So the wedge product comes in naturally via the transformation behaviour of volume. In your question it would enter on transforming the $\Delta A$ term (like in the proof of the transformation formula). In the polar coordinates example $\phi$ from the transformation formula corresponds to the coordinate change.