The Poincare disk and upper half-plane models are related by a Möbius transformation that maps the disk to the plane. Such a transformation has the form
$$
f(z) \;=\; \frac{az+b}{cz+d}
$$
for some constants $a,b,c,d\in\mathbb{C}$ such that $ad-bc\ne 0$.
In my mind, the "usual" Möbius transformation $f$ satisfies $f(-i) = 0$, $f(1) = 1$, $f(-1)=-1$, and $f(i)=\infty$, though there may be other conventions. Solving for the coefficients gives the formula for $f$:
$$
f(z) \;=\; \frac{z + i}{iz+1}.
$$
The inverse of this maps the upper half-plane to the disk, and is given by
$$
f^{-1}(z) \;=\; \frac{z-i}{-iz+1}
$$
Note For the specific requirements you gave for $f$ where $f(0) = i$, $f(-i)=\infty$, and $f(i) = 0$, you would want the $180^\circ$ rotation of the function $f$ above, i.e. $f(z) = \dfrac{z-i}{iz-1}$.
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.
Best Answer
The concept of isometry requires a Riemannian manifold, and you must be careful not to separate the manifold from its metric.
If you have a manifold $M$ with metric $g$, and a second manifold $N$ with metric $g'$, then the two manifolds are isometric if there exists a diffeomorphism $\phi: M\to N$ that preserves the metric, i.e. $g(v,w) = g'(\phi_*v, \phi_*w)$. The Poincar$\acute{\textrm{e}}$ disk is isometric to the hyperbolic plane in this sense; note that the metric of the disk is not the Euclidean metric (it is given further down in the article you've linked).
A manifold $M$ with metric $g$ can be isometrically embedded in $\mathbb{R}^n$ if there exists a submanifold $N$ of $\mathbb{R}^n$ with the induced Euclidean metric which is isometric to $M$. The hyperbolic plane cannot be isometrically embedded into $\mathbb{R}^2$, but the disk model is not a contradiction, since the metric of the disk model is not the Euclidean metric.
So the two terms do not have completely different meaning; an isometric embedding implies existence of an isometry to a submanifold in Euclidean space with one particular metric.