Why does the magnitude of the cross-product of a and b give the area of a parallelogram spanned by a and b?

Choose coordinates so that the two vectors $\vec a, \vec b$ are in the $xy$-plane, with $\vec a$ along the $x$-axis. (Note that as long as you've decided on a unit length, exactly which direction you choose for the coordinate axes doesn't change anything. The vectors and their cross product live in a coordinate-free space, just floating around. We're just imposing coordinates to make concrete calculations simpler.) That means we can set $\vec a = (a_1, 0, 0)$ and $\vec b = (b_1, b_2, 0)$. This gives $$ \vec{a}\times \vec b = (0, 0, a_1b_2) $$ and the length of this vector is $\sqrt{(a_1b_2)^2} = |a_1b_2|$, obviously. But the parallelogram has base $|a_1|$ and height $|b_2|$, which means that the area of the parallelogram is given by the exact same expression.

It is probably because the answer is simple in terms of classical 2D geometry.

$||\vec u\times \vec v||=||\vec u||.||\vec v||.\sin(\vec u,\vec v)$

But the area of the parallelogram defined by $\vec u$ and $\vec v$ is the base multiplied by the height. If you take $\vec u$ as the base, the height is $h=||\vec v||.\sin(\vec u,\vec v)$, hence the result...