Roughly speaking, the answer is "because you have a (bi)cone".
Any smooth embedded manifold (in fact any $C^2$ manifold) will have the property that the appropriately defined density is precisely equal to $1$. But if you have something that is not smooth, you can have other values.
If you have just a single cone $z = \sqrt{x^2 + y^2}$, near the origin you see that locally there is a constant angle defect: a circle of radius $r$ from the origin on the cone has circumference $\sqrt{2}\pi r < 2 \pi r$. For smooth manifolds the angle defect is related to the integral of the curvature inside the circle, so if the curvature is a continuous function (which can be guaranteed by the manifold being $C^2$), the angle defect must become zero as the circle shrinks. The single cone, on the other hand, has a curvature singularity at the origin, which allows it to have an angle defect even as the circle shrinks to a point.
Notice that a computation will show you that the single cone has density $1/\sqrt{2}$ at the origin. (The double cone, then, will have twice that.)
Remark 1 It is not necessary to have a multisheeted surfaces to get density $>1$. Consider $\mathbb{R}^3$ in spherical coordinates $(r,\phi,\theta)$. Define a surface by $\theta = f(\phi)$ with $|f(\phi)| < \pi/2$, this will by definition be a graph over $\mathbb{R}^2$ and thus single sheeted. Choose $f(\phi) = \sin (k\phi)$ for $k\in\mathbb{Z}$. The circumference of the unit circle then is given by $ \int_0^{2\pi}\sqrt{1 + k^2\cos^2(k\phi)}\cos\circ\sin(k\phi)d\phi > 2\pi $ if $|k| > 1$. So one gets that the graph defines a $C^0$ embedded surface with density $> 1$.
Remark 2 Very roughly, the density is a statement of how "crunched up" our set is in an infinitesimal setting. You should compare the notion, and the statements given in Florian's answer, to the notion of fractal dimension.