What is the geometric intuition behind $$\int_{S^1}x\,dy-y\,dx=2\pi,\qquad\int_{S^2}x\,dy\,dz-y\,dx\,dz+z\,dx\,dy=4\pi r^2,$$and in general, $$\sum_{k=1}^{n+1}(-1)^{k+1}x_k\,dx_1\cdots\widehat{dx_k}\cdots dx_{n+1}$$ being the "correct" volume form for the unit $n$-sphere standardly embedded in $\mathbf R^{n+1}$? (Yes, technically, it has to be pulled back under the inclusion map to be integrated...)
I can sorta-kinda convince myself that the 1-dimensional case works, but my (abysmal) intuition doesn't really generalize to the 2-dimensional case. In fact, I'm not even sure how to visualize the 2-form above (or any other less-than-simple 2-form, for that matter).
In general if you have a hypersurface $M^n$ inside $\mathbb R^{n+1}$, let $v$ be a unit normal vector fields on $M$, then $\iota _v( dx_1\wedge\cdots \wedge dx_{n+1})$ is a nowhere vanishing $n$-form on $M$ with constant length $1$. So
$$Vol (M) = \int_M \iota_v (dx_1\wedge \cdots \wedge dx_{n+1}).$$
In the case when $M = \mathbb S^n$ is the unit sphere, then $v(x) = (x_1, \cdots, x_{n+1})$, so
$$\iota_v (dx_1\wedge \cdots \wedge dx_{n+1}) = \sum_{k=1}^{n+1}(-1)^{k+1}x_k\,dx_1\cdots\widehat{dx_k}\cdots dx_{n+1}$$
is the "correct" volume form to integrate.