In an earlier post to math.stackexchange I asked a question beginning with:
Let $\alpha$ be the $1$-form and $\beta$ the $2$-form on $\mathbb{R}^3$ given by
$$\alpha=(x+y)\,dy+(x^2-y^2)\,dz$$
$$\beta=z\,dx\wedge dy+xz\,dx\wedge dz$$
The wedge product I got was $\alpha\wedge\beta(x,y,z)=-yz(x+y)\,dx\wedge dy\wedge dz$
My understanding was that $\alpha$ would be in terms of one of the components in $\mathbb{R}^3$ and $\beta$ would be in terms of two of the components in $\mathbb{R}^3$, but this is not the case at all.
It made sense for $\alpha\wedge\beta$ to be a three form to me since it is in terms of $x,y$ and $z$
But this did not hold for $\alpha$ or $\beta$, so this cannot be the relationship.
One other thing I noticed was that $\alpha$ has just single $dy$ and $dz$ terms, whereas $\beta$ has two differential forms wedged, and $\alpha\wedge\beta$ the three form has three differential forms wedged, is this the relationship?
It makes sense to me for this to be the relationship, but I cannot find a source to verify this for me.
In $\Bbb R^3,$ every $1-$form writes as some combination $$\alpha = f \ {\rm d}x + g \ {\rm d}y + h \ {\rm d}z,$$ $2-$forms writes as combinations $$\beta = f \ {\rm d}x \wedge {\rm d}y+ g \ {\rm d}x \wedge {\rm d}z+ h \ {\rm d}y\wedge {\rm d}z,$$ and $3-$forms are $$\omega = f \ {\rm d}x \wedge {\rm d}y\wedge {\rm d}z.$$ You can find proofs about this in Do Carmo's "Differential Forms and Applications", and a little in O'Neill's Elementary Differential Geometry. The wedge product of a $k-$ form with a $s-$form is a $(k+s)-$form. Since the wedge product is anti-commutative, repeats kill. So in your example: $$\alpha \wedge \beta =(x+y)xz \ {\rm dy} \wedge {\rm d}x \wedge {\rm d}z + (x^2-y^2)z \ {\rm d}z \wedge {\rm d}x \wedge {\rm d}y \\ = (x^2z-y^2z-x^2z-xyz) \ {\rm d}x \wedge {\rm d}y \wedge {\rm d}z = -(y^2z+xyz) \ {\rm d}x \wedge {\rm d}y \wedge {\rm d}z, $$ so your result is correct.