I've managed to confuse myself with cones and deficit angles. Let's consider a conical defect in 2 dimensions. So the metric is the usual one in polar coordinates, $$ ds^2 = dr^2 + r^2 d\phi^2,$$ except that now $\phi \sim \phi + 2\pi(1-\alpha)$. For $\alpha = 0$, the is just flat space. When $\alpha \neq 0$, there is a singularity in the curvature, and for example the Ricci scalar acquires a delta function: $$R(x) = 4\pi \alpha \, \delta^{(2)}_{x,x'}, $$ where $x'$ is the location of the conical defect. Now if we use the Gauss-Bonnet theorem (remembering that in 2 dimensions $R=K/2$, where $K$ is the Gaussian curvature), we can relate the deficit angle to the Euler characteristic (neglecting any boundary terms) $$ \chi = \alpha.$$
So my confusion now is: what does it mean to have $\alpha = 1$, which is to say that the deficit angle is $2\pi$? It seems weird that I can remove the whole angle and still have a 2 dimensional space. Since I don't have much intuition for what it means to remove $2\pi$, I looked up what manifolds have $\chi = 1$, I find things like the disk (which has a boundary), and the real projective plane, which is $S^2/\mathbb{Z}_2$ and non-orientable.
So what space is a cone with deficit angle $2\pi$? (Bonus question: what space has deficit angle $4\pi$, the Euler formula would suggest a sphere?)
The "cone of zero angle" is a cusp. Unlike for positive angles, there is no single canonical model of a cusp: they come in various shapes. For example, the surface $z=(x^2+y^2)^{p}$ where $p\in(0,1/2)$ has a cusp at the origin. The choice $p=1/4$ results in a nice algebraic surface $x^2+y^2=z^4$ in two symmetric parts joined by a cusp.
One does not get a cusp by taking a quotient of the plane. The surfaces mentioned above are not flat outside of the cusp, and this curvature contributes to the Gauss-Bonnet formula. So does the boundary term. The surface $z=(x^2+y^2)^{p}$ is topologically a disk.
The conclusion $\chi=\alpha$ in your post is not correct; the boundary term can't be simply ignored. The cones for various $\alpha\in(0,1)$ are mutually homeomorphic.