When Gauss-Bonnet thm is used on a constant Gauss curvature surfaces of $ (~ K=-1/a^2,K=0,K<=+1/a^2)$ along a closed radial contours, the exterior angle rotation sum can be shown to be respectively:
$$ \Sigma \psi= 2 \pi (1+\Delta r/a),~~2 \pi (1+ 0\cdot r/a),~~2 \pi (1+\Delta z/a) $$ Here $(\Delta z,\Delta r)$ represent axial or radial difference of truncated shell surface area considered for $ (K>0, K<0) $ cases (I had also shown this before on this site).
For the central Pseudosphere aka Tracticoid we have area computation
$$ dA=2 \pi r\cdot ds= 2 \pi r \cdot {dr}/ \sin \phi= 2 \pi a \cdot dr $$
$$ A=2 \pi \int_{r_1}^{r_2} dr = 2 \pi a \cdot \Delta r$$
It can be shown by integration that area of each Nappe or segment is $4 \pi a^2 $ for non central cases as well that also verifies Isometry.
Considering Monkey saddle or Pringle's chip topography, the angle sum $ >2 \pi$ can appeal to intuition.
But at the pseudo-sphere maximum radius $(r=a)$ the angle sum of two full turnings at cuspidal edge as exactly $ 4 \pi$ is still open to interpretation.. (to me now) is counter-intuitive.
Is the tally of $2 \pi$ for each of two horns / Nappes?
Thanks for help in interpreting this definite limit when encountering the edge of singularity.
I think you are mis-applying the Gauss-Bonnet theorem. It says
$$\oint k_g\,ds+\iint K_G\,dS=2\pi\,\chi$$
where $S$ is surface area of a region of the surface, $K_G$ is its Gaussian curvature, $s$ is arclength along the region's boundary, $k_g$ is its geodesic curvature (positive inward, negative outward), and $\chi$ is the region's Euler characteristic. $\oint k_g\,ds$ is equivalent to your $\Sigma\psi$.
If the $x$-axis is the tractricoid's axis of symmetry (going left to right in the wiki image), and $y=r\cos\theta,\;z=r\sin\theta$, then the area of a region $0<x_2<x<x_1$ is indeed
$$S=2\pi a(r_2-r_1).$$
But this region is not a disk! It's topologically an annulus, with Euler characteristic $\chi=0$, not $1$, so the Gauss-Bonnet theorem gives
$$\Sigma\psi=-\iint(-1/a^2)\,dS+2\pi\,0$$
$$=S/a^2=2\pi(r_2-r_1)/a.$$
The right half of the tractricoid has $x_2\to0,\;x_1\to\infty$, or equivalently $r_2\to a,\;r_1\to0$, so the area is $2\pi a^2$.
$\Sigma\psi=\Sigma\psi_2+\Sigma\psi_1$ involves the two circles at $x_2$ and $x_1$; but the circle on the right has constant geodesic curvature $-1/a$ regardless of $x_1$, so $\Sigma\psi_1=\int(-1/a)\,ds=-2\pi r_1/a$, and likewise the circle on the left has $\Sigma\psi_2=2\pi r_2/a$, so in fact we've just verified the theorem in this case.
Anyway, as $r_2\to a$, we have $\Sigma\psi_2\to2\pi$. This is to be expected, because the tractricoid's tangent plane approaches being vertical, that is, parallel to the circle's extrinsic curvature vector; so the geodesic curvature is the same as the extrinsic curvature.