Let's say we have some 4D maniforld $M$ in which the differential volume is given by Hodge's star $\star$ as
$$ dV :=\star\ 1 = \sqrt{|g|}\ dx^0\wedge dx^1\wedge dx^2\wedge dx^3 $$
Where $g$ is the determinant of the metric tensor on $M$
Now, this $dV$ should be equal to the usually $dv$ given as
$$ dv = \sqrt{|g|}\ dx^0dx^1dx^2dx^3 $$
since both of them are the same concept, represent the same quantity. But since the wedge product is define as
$$ dx\wedge dy = \frac{1}{2}(dx\otimes dy - dy\otimes dx) $$
I don't see how $dV$ and $dv$ are equal and moreover I don't understand quite well the importance of $\otimes$ in the wedge's definition. What would be the difference betweeen integrating a function $f(x, y)$ with $dx\otimes dy$ instead of $dy\otimes dx$? And how to prove $dV = dv$?