the Riemannian volume form
Once a manifold carries a metric, you know how long vectors are and what angle they make, so you ought to be able to say how much n-dimensional volume sits in a little parallelepiped — and to integrate functions over the manifold. The Riemannian volume form is exactly the n-form that packages 'metric volume' so that calculus on M has an honest notion of total mass, area, or volume.
On an oriented Riemannian n-manifold (M, g) the volume form dV_g is the unique positive n-form that gives the value 1 to any positively oriented orthonormal frame e_1, ..., e_n. In local oriented coordinates it is dV_g = sqrt(det(g_ij)) dx^1 ^ ... ^ dx^n, where det(g_ij) is the determinant of the matrix of metric coefficients. The factor sqrt(det g) is exactly the volume of the parallelepiped spanned by the coordinate vectors d/dx^i, which is why a coordinate box of side dx^i holds sqrt(det g) times the naive coordinate volume. The total volume of M is the integral over M of dV_g.
Two cautions. If M is not orientable there is no global volume FORM, but there is still a volume DENSITY |dV_g| and a well-defined measure for integrating functions, so 'volume' survives even without orientation. And dV_g depends on g: rescaling the metric by a factor c (g -> c g) multiplies the volume form by c^(n/2), so changing the metric genuinely changes volumes — there is no canonical volume on a bare smooth manifold.
On the round 2-sphere of radius r, in spherical coordinates the metric is ds^2 = r^2 dtheta^2 + r^2 sin^2(theta) dphi^2, so sqrt(det g) = r^2 sin(theta) and dV_g = r^2 sin(theta) dtheta dphi; integrating this over theta in [0, pi], phi in [0, 2pi] gives the familiar total area 4 pi r^2.
The sqrt(det g) factor is what turns a naive coordinate integral into a true geometric area.
A volume FORM needs orientation; a volume DENSITY (or measure) does not. Don't claim a non-orientable manifold has no notion of volume — it has the measure, just not the global form.