Let's first concentrate on a single line. I.e. your hyperbolic 1-space is modeled by the open interval $(-1,1)$. You have $A=-1,B=1$. Take two points in the Poincaré model, and compute the cross ratio
$$
(A,B;Q,P) =
\frac{\lvert QA\rvert\cdot\lvert BP\rvert}{\lvert PA\rvert\cdot\lvert BQ\rvert}
= \frac{(1+Q)(1-P)}{(1+P)(1-Q)}
$$
Now transfer these points into the Klein model, at your discretion either via stereographic projection and the hemisphere model, or via the hyperboloid model, or purely algebraically. You obtain
$$
P' = \frac{2P}{1+P^2} \qquad
Q' = \frac{2Q}{1+Q^2}
$$
Plug these into the cross ratio and you get
$$
(A,B;Q',P')=\frac{(1+Q^2+2Q)(1+P^2-2P)}{(1+P^2+2P)(1+Q^2-2Q)}=
\frac{(1+Q)^2(1-P)^2}{(1+P)^2(1-Q)^2} =
(A,B;Q,P)^2
$$
So the Kleinian cross ratio is the square of that from the Poincaré model. Therefore the distances will differ by a factor of two. Since cross ratios are invariant under projective transformations (of $\mathbb{RP}^2$ for Klein resp. $\mathbb{CP}^1$ for Poincaré), the above considerations hold for the plane as well.
So which coefficient is the correct one? That depends on your curvature. If you want curvature $-1$, or in other words, if you want an ideal triangle to have area $\pi$ so that angle deficit equals area, then the $\frac12$ in front of the Klein formula is correct as far as I recall. For Poincaré you'd better use coefficient $1$, then the lengths in the two models will match.
If you use coefficient $\frac12$ in the Poincaré model, then you effectively double your unit of length. All length measurements get divided by two, including the imaginary radius of your surface. Since Gaussian curvature is the product of two inverse radii, you get four times the curvature, namely $-4$, just as Post No Bulls indicated.
For the lengths in the Poincare disk models: If the hyperbolic line is an euclidean circle are the euclidean lengths measured as the segment-lengths or as arc-lengths (along the circle)?
Segment lengths (i.e. chord lengths) are certainly correct. I think of the cross ratio as one of four numbers in $\mathbb C$. If you write your differences like this
$$z_{QA}=Q-A=r_{QA}\,e^{i\varphi_{QA}}=\lvert QA\rvert\,e^{i\varphi_{QA}}
\in\mathbb C$$
then the cross ratio becomes
$$
(A,B;Q,P)=\frac{(Q-A)(B-P)}{(P-A)(B-Q)}=
\frac{r_{QA}\,e^{i\varphi_{QA}}\cdot r_{BP}\,e^{i\varphi_{BP}}}
{r_{PA}\,e^{i\varphi_{PA}}\cdot r_{BQ}\,e^{i\varphi_{BQ}}}=\\
=\frac{r_{QA}\cdot r_{BP}}{r_{PA}\cdot r_{BQ}}\,
e^{i(\varphi_{QA}+\varphi_{BP}-\varphi_{PA}-\varphi_{BQ})}=
\frac{\lvert QA\rvert\cdot\lvert BP\rvert}{\lvert PA\rvert\cdot\lvert BQ\rvert}
\in\mathbb R
$$
This is because the phases have to cancel out: the cross ratio of four cocircular points in $\mathbb C$ is a real number, so $\varphi_{AQ}+\varphi_{BP}-\varphi_{PA}-\varphi_{BQ}$ has to be a multiple of $\pi$, and in fact I'm sure it will be a multiple of $2\pi$.
This doesn't neccessarily rule out arc lengths, but a simple example using arbitrarily chosen numbers shows that arc lengths result in a different value, so these are not an option.
You do have to use circle arcs instead of chords if you compute lengths as an integral along some geodesic path. So be sure not to mix these two approaches.
The quick way to get the correct answer is dimensional analysis: to get an area from a dimensionless angle defect you need to multiply by something with units length$^2$, so Wikipedia's formula $A = (\pi - \sum\theta)R^2$ is correct.
Sommerville is a very old text and uses $k$ for the negative inverse curvature - see e.g. page 75 where it states "The formulae of hyperbolic trigonometry become of those of euclidean plane geometry as $k \to \infty$."
The proof is going to depend on how you are constructing things - from the Riemannian perspective you just calculate how curvatures and areas behave under multiplication of the metric by a constant: covariant derivatives are unchanged, so Ricci curvature is unchanged, so scalar curvature scales as the inverse metric. Area scales as the metric. Thus area is inversely proportional to curvature under scaling.
Best Answer
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$tl; dr:
There's a hyperboloid model of the projective plane that is easily shown (using elementary linear algebra, see sketch below) to be homogeneous and isotropic under its isometry group. That's enough to guarantee its Gaussian curvature is constant. (In fact, every homogeneous surface has constant Gaussian curvature; isotropy isn't needed.)
No, the curvature of a hyperbolic metric is independent of the Euclidean radius of a disk. For instance, if $a > 0$, the metric with components $$ g_{11} = g_{22} = \frac{4a^{2}}{\bigl(1 - (x^{2} + y^{2})\bigr)^{2}},\quad g_{12} = 0 $$ in the unit disk has Gaussian curvature $-a$.
In a nutshell, let $$ H = \{(x, y, z) : -x^{2} - y^{2} + z^{2} = 1, z > 0\} \subset \Reals^{2, 1} $$ be the set of unit future-pointing timelike vectors in Minkowski space, and let $g$ be the Riemannian (!) metric obtained by restricting the inner product $$ \Brak{(u_{1}, v_{1}, w_{1}), (u_{2}, v_{2}, w_{2})} = u_{1} u_{2} + v_{1} v_{2} - w_{1} w_{2}. \tag{1} $$ For all real $\theta$ and $\alpha$, the set \begin{align*} e_{1} &= (\cosh\alpha \cos\theta, \cosh\alpha \sin\theta, \sinh\alpha), \\ e_{2} &= (-\sin\theta, \cos\theta, 0), \\ e_{3} &= (\sinh\alpha \cos\theta, \sinh\alpha \sin\theta, \cosh\alpha), \end{align*} is orthonormal with respect to the inner product (1). The first two vectors are spacelike, and the last is timelike (in fact, an element of $H$).
The general positive orthonormal basis is obtained by "rotating the $(e_{1}, e_{2})$-plane": $$ (\cos\phi) e_{1} + (\sin\phi) e_{2},\quad (-\sin\phi) e_{1} + (\cos\phi) e_{2},\quad e_{3}. \tag{2} $$
The families of linear transformations with standard matrices $$ \left[\begin{array}{@{}ccc@{}} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \phantom{-}\cos\theta & 0 \\ 0 & 0 & 1 \end{array}\right],\qquad \left[\begin{array}{@{}ccc@{}} \cosh\alpha & 0 & \sinh\alpha \\ 0 & 1 & 0 \\ \sinh\alpha & 0 & \cosh\alpha \\ \end{array}\right], $$ namely rotations about the $z$-axis and boosts in the $(x, z)$-plane, act transitively on the set of positive orthonormal bases, and consequently generate the isometry group of $(H, g)$. (Each preserves both $H$ and the Minkowski inner product, hence induces an isometry. To show the action is transitive, start with an arbitrary positive orthonormal basis; rotate about the $z$-axis so that the timelike vector lies in the $(x, z)$-halfplane with $x > 0$; boost to make the third vector $(0, 0, 1)$; rotate about the $z$-axis to make the spacelike vectors $(1, 0, 0)$ and $(0, 1, 0)$.)
To show the curvature of $(H, g)$ is $-1$, it suffices to calculate at the point $(0, 0, 1)$. (Left to you.)
Hyperbolic lines are precisely the intersections of $H$ with a plane through the origin. (The intersection of $H$ with a plane containing the $z$-axis is the fixed-point set of a reflection isometry, hence a geodesic of $(H, g)$. Every plane through the origin that intersects $H$ can be brought to a plane containing the $z$-axis by a boost. Given points $p$ and $q$ in $H$, there is a unique plane containing $p$, $q$, and the origin; the intersection of this plane with $H$ is the geodesic from $p$ to $q$.)
The Klein model is the open unit disk in the plane $\{z = 1\}$, a.k.a., the image of $H$ under radial projection from $(0, 0, 0)$. A line is this model is a chord of the unit circle.
The Poincaré model is the open unit disk in the plane $\{z = 0\}$, a.k.a., the image of $H$ under radial projection from $(0, 0, -1)$.