Gauss Bonnet theorem validation with hyperbolic circles How to verify the theorem in case of a hyperbolic circle radius $r$ for constant negative Gauss curvature $ K=-1/a^2 $ and constant geodesic curvature $k_g $ ...
 e.g., like here...
$$k_g=1/r \tag1 $$
$$  \int k_g ds + \int\int K dA = 2 \pi \tag2 $$
(Geodesic polar coordinates) Perimeter and Area plugged in from : 
$$ Perimeter = 2 \pi a \sinh (r/a) ,\, Area= 4 \pi a^2 \sinh^2(r/2a) \tag3 $$
as these reduce to $ ( 2\pi r, \pi r^2) $ when $ a\rightarrow \infty$
$$\frac{\pi a \sinh (r/a)   }{ r } -{ 4 K \pi a^2 \sinh^2 (r/2a)} = 2 \pi $$
does not tally in general. Also when $ r\rightarrow 0,  \pi + \pi (r/a)^2 \ne 2 \pi$
Clearly (1) is assumed intentionally wrong, but then what is correct ? It needs to be defined  properly in the tangent plane or in the hyperbolic plane which is not known to me.
Since a direct calculation (like for case $ K=+1/a^2 $)  appears elusive.. an indirect back -calculation is the only resort . For this case we have after some simplification:
$$  \boxed{a \, k_g= \coth (r/a ), R_g=a \tanh (r/a)}\tag4$$
verified by plugging this into GB thm (1). When 
$$r\rightarrow \infty,\,k_g \rightarrow \,(1/a),\, R_g \rightarrow \,a \tag5 $$ 
$$r=0 ,k_{g} = \infty \tag6 $$ 
as expected for a point circle.
(4) is a new result (for me) surprising and apparently anomalous because it is bounded. There is no way to check it directly or conceptually reinforcing it by other theorems or better known results in hyperbolic geometry. 
And that is the motivation of this posting... 
Show how theses hyperbolic circles are/can be placed geometrically in the hyperbolic models available (Poincaré half plane, Poincaré disc and Klein's )
 A: Fix $a > 0$. As you've found, in the hyperbolic plane with curvature $-\frac{1}{a^{2}}$, a circle of hyperbolic radius $r$ has geodesic curvature
$$
k_{g} = \tfrac{1}{a} \coth \tfrac{r}{a}.
\tag{*}
$$
The integral is
$$
\int k_{g}\, ds = k_{g} \times \text{perimeter} = 2\pi \cosh \tfrac{r}{a};
$$
since the enclosed total Gaussian curvature is
$$
4\pi \sinh^{2}\tfrac{r}{2a} = 2\pi(\cosh \tfrac{r}{a} - 1),
$$
the Gauss-Bonnet theorem yields the expected result.

Here are a couple of ways to see why (*) is true.
First, the geodesic curvature of a circle of intrinsic radius $r$ on a sphere of radius $a$ is $\frac{1}{a} \cot \frac{r}{a}$. On the principle that "corresponding" spherical and hyperbolic quantities are given by corresponding formulas with circular functions replaced by hyperbolic functions, we'd expect the geodesic curvature of a hyperbolic circle of radius $r$ to be $\frac{1}{a} \coth \frac{r}{a}$.
Second (and less magically), model the hyperbolic plane as the open unit disk in the Cartesian plane, equipped with the metric
$$
g = \frac{4a^{2}(dx^{2} + dy^{2})}{(1 - (x^{2} + y^{2}))^{2}}.
$$
If $0 \leq R < 1$, the radial segment $(t, 0)$ with $0 \leq t \leq R$ has hyperbolic length
$$
r := \int_{0}^{R} \frac{2\, dt}{a^{2} - t^{2}}
= a\log \frac{a + R}{a - R};
$$
that is, the circle of hyperbolic radius $r$ centered at the origin has Euclidean radius
$$
R = \tanh \tfrac{ar}{2},
\tag{1}
$$
and hyperbolic circumference
$$
\frac{2a(2\pi R)}{1 - R^{2}} = 2\pi a \sinh \tfrac{r}{a}.
$$
To find the geodesic curvature of a circle $C$ of hyperbolic radius $r$, draw the circle of Euclidean radius $R$ centered at the origin, fix a point $p$ of this circle, and draw the hyperbolic line $\ell$ tangent to the circle at $p$. Because $\ell$ is the arc of a Euclidean circle crossing the unit circle at right angles, a bit of geometry shows $\ell$ is part of the Euclidean circle of center $(\frac{1 + R^{2}}{2R}, 0)$ and radius $R' := \frac{1 - R^{2}}{2R}$.

Let $d\theta$ be a small angle at the center of $C$, subtending an arc of hyperbolic length
$$
ds = \frac{2aR}{1 - R^{2}}\, d\theta = a(\sinh \tfrac{r}{a})\, d\theta.
\tag{2}
$$
If $d\phi$ is the indicated Euclidean angle subtended at the Euclidean center of the circle representing $\ell$, then (up to first order in $ds$)
$$
R\, d\theta = R'\, d\phi,\qquad\text{i.e.,}\quad
\frac{d\phi}{d\theta} = \frac{R}{R'} = \frac{2R^{2}}{1 - R^{2}}.
$$
By (1), we have
$$
1 + \frac{d\phi}{d\theta}
= \frac{1 + R^{2}}{1 - R^{2}}
= \frac{1 + \tanh^{2} \frac{r}{2a}}{1 - \tanh^{2} \frac{r}{2a}}
= \cosh \tfrac{r}{a}.
\tag{3}
$$
Parallel transport along $C$ of the boldface tangent vector at $p$ is the vector labeled $v$. The relative angle of turning of the tangent $v'$ to the circle is $d\alpha = d\theta + d\phi$. The geodesic curvature at $p$ is the rate at which the tangent to $C$ rotates per unit length. By (2) and (3), this is equal to
$$
  k_{g} = \frac{d\alpha}{ds}
  = \frac{d\theta + d\phi}{ds} = \left(1 + \frac{d\phi}{d\theta}\right) \frac{d\theta}{ds}
  = \frac{\cosh \frac{r}{a}}{a \sinh \frac{r}{a}} = \tfrac{1}{a} \coth \tfrac{r}{a}.
$$
