The connection here is the Minkowski space, which can be used to describe both.
Hyperbolic geometry
For example, take hyperbolic 2-space in the hyperboloid model. You'd represent hyperbolic points as points on the hyperboloid, namely as
$$\left\{(x_1,x_2,x_3)^T\;\middle|\;x_1^2-x_2^2-x_3^2=1\right\}$$
This expression $x_1^2-x_2^2-x_3^2$ is the quadratic form which lies at the foundation of the Minkowski space $\mathbb R^{1,2}$. The corresponding bilinear form can be used to compute distances, as
$$d(x,y)=\operatorname{arcosh}\left(x_1y_1-x_2y_2-x_3y_3\right)$$
You can even think about this projectively: you may use a vector which does not lie on the hyperboloid, then use that vector to define a line which will intersect the hyperboloid in a given point, which is the point it specifies. This will work for every vector whose quadratic form is positive.
This idea is also very useful to define lines. A hyperbolic line (i.e. a geodesic) connecting two hyperbolic points is modeled by the intersection between the hyperboloid and a plane spanned by these two points and the origin. You can describe this plane by its normal vector, and you can compute that normal vector as the cross product of two vectors representing the two points. Conversely, you can obtain the intersection between two geodesics by computing the cross product between two normal vectors of such planes, although the quadratic form for that point likely won't be $1$ yet, but any other positive value instead. Therefore, this normal vector of the plane is a reasonable (and homogenous) representation of the line. Its quadratic form will be negative.
Common vocabulary
Now change the vocabulary to use terms which are common for Minkowski spaces. A vector whose quadratic form is positive is said to be time-like. So points of the hyperbolic plane correspond to time-like vectors, with scalar multiples of a vector representing the same point. Likewise a vector whose quadratic form is negative is called space-like. So a line in hyperbolic geometry corresponds to a space-like vector, and all its multiples. In between these two, there are those vectors for which the quadratic form is zero. These correspond to ideal points of your geometry. in a certain sense, an ideal point is as much a line as it is a point.
The set of hyperbolic isometries are those linear transformations of your vector space which preserve the set of ideal points, i.e. which preserve the light cone. These correspond roughly to Lorentz transformations in relativistic vocabulary (with some care because here we identify scalar multiples but there we don't, but the central idea of preserving the light cone remains).
Relativistic geometry
So where do these physically sounding terms come from? Imagining the whole vector space as some kind of space-time-diagram should be fairly simple. The first dimension (with the positive sign) would be time, the other two would be space. A vector would denote an event in this diagram. An event where all spatial coordinates are zero would happen at the same place as the origin, but at some different time. An event with zero time coordinate would happen at the same time but at some other place. The light cone would correspond to a cone of slope $1$, which is the speed of light in our coordinate system. Light travels from the origin along the light cone.
But time and space are relative, so the above choice of coordinate system is only valid for a given inertial system. To convert between inertial systems which meet at the origin, you'd again use a Lorentz transformation, i.e. a transformation which preserves the light cone. Using such a transformation, any event which is a time-like distance away from the origin can be made to happen at the same place but in the past or the future. You'd use an inertial system which travels to that event or came from it. Likewise, any event which is a space-like distance away can be made to happen at the same time, using the right movement to compensate.
Conclusion
So conversions between inertial systems correspond to isometric transformations of the hyperbolic space. And objects in hyperbolic space correspond to (equivalence classes of) events in space-time diagrams.
The above would generalize for higher dimensions, but the part about two points spanning a line would be more complicated to read, since you'd more likely talk about three points spanning a plane. $ $ $ $
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.
Best Answer
You asked
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.