Here's another version of Doug Chatham's answer, but with details.
If you lived in Hyperbolic space, then Euclidean geometry would be natural to you as well. The reason is that you can take what is called a horosphere (in the half-space model for us, this is just a hyperplane which is parallel to our limiting hyperplane) and this surface actually has a Euclidean geometry on it!
So unlike for us, where the hyperbolic plane cannot be embedded into Euclidean 3-space, the opposite is true: the Euclidean plane can be embedded into hyperbolic 3-space! So this is analogous to our understanding of spherical geometry. It's no surprise the spherical geometry is slightly different, however, it fits nicely into our Euclidean view of things, because spherical geometry is somewhat contained in three-dimensional geometry because of the embedding.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$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 $(0, 0, -1)$.
Best Answer
In differential geometry, a hyperbolic plane is a complete, simply-connected Riemannian $2$-manifold equipped with a metric of constant negative (Gaussian) curvature. By scaling the metric, we may assume the curvature is $-1$; often when people say hyperbolic plane they're assuming $K \equiv -1$. The rest of this answer assumes $K \equiv -1$.
It turns out that any two hyperbolic planes of curvature $-1$ are isometric as Riemannian manifolds. In my experience, when people speak of the hyperbolic plane, they refer to the equivalence class of hyperbolic planes in the space of Riemannian manifolds up to isometry.
It's possible that a hyperbolic model in the sense of the question is what I've called a hyperbolic plane above. There are a number of standard models. The first three below naturally sit inside Minkowski three-space, the real Cartesian three-space equipped with the metric $dx^{2} + dy^{2} - dz^{2}$.
The hyperboloid model, the upper sheet of a two-sheeted hyperboloid, $x^{2} + y^{2} - z^{2} = -1$, $z > 0$. Geodesics ("hyperbolic lines") in this model are intersections of the hyperboloid with planes through the origin. (This surface has non-constant _positive curvature with respect to the Euclidean ambient metric, but acquires constant negative curvature from the ambient Minkowski metric.)
The conformal disk model (or Poincaré model), the open unit disk in the plane $z = 0$ with the metric $\frac{4(dx^{2} + dy^{2})}{(1 - (x^{2} + y^{2}))^{2}}$ induced by projection from the point $(0, 0, -1)$, see diagram. Because the metric is a scalar multiple of the Euclidean metric $dx^{2} + dy^{2}$, hyperbolic angles coincide with Euclidean angles. Geodesics turn out to be arcs of circles meeting the unit circle at right angles. At first glance this metric appears flat; see below.
The affine disk model (or Klein-Beltrami model), the open unit disk in the plane $z = 1$ with the metric induced by projection from the origin. Geodesics in this model are Euclidean segments. If the open unit disk and projection point are translated one unit up the $z$-axis in the preceding model, we obtain the affine model.
The conformal upper half plane model, induced from the conformal disk by a fractional linear transformation. The metric is $\frac{dx^{2} + dy^{2}}{y^{2}}$, and geodesics are vertical lines and semicircles centered on the $x$-axis.
Curvature is a measure of angular defect per unit area in geodesic triangles. Qualitatively, a surface has negative curvature if the total interior angle of a geodesic triangle (enclosing a topological disk, which is automatic in the hyperbolic plane) is smaller than $\pi$. Given that hyperbolic and Euclidean angles coincide in the conformal models, and that geodesics are arcs of circles, it's visually plausible that the conformal disk and half-plane models have negative curvature.
If we represent a region of the hyperbolic plane as a surface in Euclidean three-space, we obtain a surface having "saddle-like" geometry at each point. The best-known examples are surfaces of rotation, especially the pseudosphere, and surfaces with helicoidal symmetry such as Dini's surface.
A hyperbolic circle of hyperbolic radius has hyperbolic circumference $2\pi \sinh r = \pi(e^{r} - e^{-r})$. Particularly, circumference grows exponentially with radius. The theorem of Hilbert mentioned asserts that the entire hyperbolic plane cannot be represented, even allowing self-intersection. Qualitatively (thinking of crocheted models that start like a saddle and grow outward radially), a hyperbolic disk is "floppy", and the larger the radius, the more area must be accommodated in a Euclidean ball whose Euclidean radius grows like the hyperbolic radius of the disk; "Euclidean space just can't keep up".
As noted in the comments, however, a hyperbolic disk of arbitrarily large hyperbolic radius can be isometrically embedded in Euclidean three-space. One method is to use Dini's surface, taking the edge of the disk to lie on the edge of the surface, and taking the center to lie "exponentially far away along the horn", so that the disk is tightly wrapped around the "axis".
Coda: Hyperbolic geometry also overlaps synthetic geometry. Euclid's parallel postulate is replaced by another, equivalent to "Given a line $\ell$ and a point $p$ not on $\ell$, there exist two lines through $p$ that do not meet $\ell$." Pat Ryan's Euclidean and Non-Euclidean Geometry is an accessible treatment, including the hyperboloid model and the necessary geometry of Minkowski space, and requiring little more than one semester of linear algebra.