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.
You asked
Because of this, is there a model of hyperbolic geometry that resolves all of these "problems" (i.e. it uses Euclidean metric, a negatively-curved surface, simple geodesics/lines, and a familiar geometric framework), or is it impossible?
and the simple answer is it is not possible. The reason is that
what you want can only be done for a geometry that has greater
curvature than the space it is embedded in. Thus, in Euclidean
space of zero curvature we can have nice models of positively
curved surfaces. If we lived in a negatively curved space, then
we could have nice models of surfaces of greater curvature such
as the Euclidean plane. However, we still could not have nice
models of surfaces with more negative curvatures.
One reason, among others, for this situation is that
the perimeter of a circle expands linearly as the radius
increases in Euclidean spaces. In hyperbolic spaces,
the perimeter expands essentially exponentially. Thus,
there is no room in Euclidean space to contain all of
the perimeter in a nice way without compromises.
Best Answer
As others have indicated one has to distinguish a metric $(x,y)\mapsto d(x,y)$ which measures distances between points $x$, $y$ in a space $X$ and a Riemannian metric which tells us how lengths of curves $\gamma$ in a manifold $X$ should be computed.
In the complex plane we have the usual euclidean metric $d(z_1,z_2):=|z_2-z_1|$ and at the infinitesimal scale the Riemannian metric $$ds^2:=|dz|^2=dx^2+dy^2\ .$$ The latter formula says that the length of an arbitrary curve $$\gamma: \quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)$$ should be computed as $$L(\gamma)=\int_\gamma ds=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\ dt = \int_a^b|z'(t)|\ dt\ .$$ This formula implies that the length of a segment $\sigma$ connecting two points $z_1$ and $z_2$ is just $|z_2-z_1|$.
On the "Poincare disc" $P$ we are given a priori only a Riemannian metric $$ds:= {|dz|\over 1-|z|^2}$$ (resp., $ds^2=\ldots$). This metric allows us to compute the lengths of arbitrary curves in $P$: $$L(\gamma)=\int_a^b{|z'(t)|\over1-|z(t)|^2}\ dt\ .$$ The particular definition of $ds$ is chosen such that this hyperbolic length is invariant under arbitrary conformal movements of $P$ and the curves therein.
A posteriori one can define a metric $d(\cdot,\cdot)$ on $P$ by letting the distance $d(z_1,z_2)$ between two points $z_1$, $z_2\in P$ be the hyperbolic length of the shortest curve $\gamma$ connecting $z_1$ and $z_2$. The actual carrying out of this idea shows that $d(z_1,z_2)$ can be written as an elementary function (using ${\rm artanh}$, etc.) in terms of $z_1$ and $z_2$.