I define the hyperbolic plane as the space $$\mathbb{H} = \left\{ (x,y) \in \mathbb{R}^2 | y>0 \right\}$$ endowed with the metric $$\mathrm{d}s^2_{\mathbb{H}} = \frac{\mathrm{d} x^2 + \mathrm{d} y^2}{y^2} \, .$$ I want to study here various "models" of this hyperbolic plane. Most of what I will say has an analogous un higher dimension, but I will keep the discussion as simple and as visual as possible, so I will stick to two dimensions.

### The Poincaré disk

First of all, I recall that there is a version of the hyperbolic plane that has the shape of a disk. To see that, define the complex coordinate $z=x+iy$, and define the new complex coordinate $u$ by the relation $$iu = \frac{1+iz}{1-iz} \qquad \Leftrightarrow \qquad iz = \frac{-1+iu}{1+iu} \, .$$ It is easy to check that this establishes a bijection between $\mathbb{H}$ parametrized by $z$ and the Poincaré disk $$\mathbb{D} = \{ u \in \mathbb{C} \mid |u|<1 \} \, .$$ The metric on the hyperbolic plane and the Poincaré disk in complex coordinates are \begin{equation} \label{met1} \mathrm{d} s^2_{\mathbb{H}} = \frac{\mathrm{d} z \mathrm{d} \bar{z}}{(\mathrm{Im} \, z)^2} = \frac{4 \mathrm{d} u \mathrm{d} \bar{u}}{(1-|u|^2)^2} \, . \end{equation} One can check that the Ricci scalar is constant and equal to $R=-2$ (obviously, this can be computed in any of the three models).

### Embedding in Minkowski space

Now consider the upper sheet of a hyperboloid in Minkowski space $\mathbb{R}^{1,2}$. The metric on this space is $\mathrm{d} s^2_{\textrm{Mink}} = -\mathrm{d} X_0^2 + \mathrm{d} X_1^2 + \mathrm{d} X_2^2$, and the hyperboloid sheet is defined by the equation $$X_0^2 - X_1^2 - X_2^2 = 1 \, , \qquad X_0>0 \, .$$ We can parametrize this by two angles $\alpha$ and $\beta$: \begin{eqnarray*} X_0 &=& \cosh \alpha \\ X_1 &=& \sinh \alpha \cos \beta \\ X_2 &=& \sinh \alpha \sin \beta \end{eqnarray*} Using these coordinates, we find that the metric on the sheet is \begin{equation} \label{met2} \mathrm{d} s^2_{\mathbb{H}} = \mathrm{d} \alpha ^2 + (\sinh \alpha)^2 \mathrm{d} \beta^2 \, , \end{equation} and again we can compute the Ricci scalar and find $R=-2$. The relation between this model and the Poincaré disk model is given by a stereographic projection. More precisely, the Poincaré disk is the stereographic projection of the hyperboloid sheet on the plane $X_0=0$ with respect to the point $(X_0,X_1,X_2)=(-1,0,0)$. Let us work out the details: a point $(X_0,X_1,X_2)$ on the hyperboloid sheet is mapped by this projection to the point $$\left( 0 , \frac{X_1}{1+X_0} , \frac{X_2}{1+X_0} \right) \, .$$ Using that, we obtain the relation between the coordinate on the disk $u$ and the coordinates $\alpha$ and $\beta$ on the hyperboloid, $$u = \frac{\sinh \alpha}{1+\cosh \alpha} e^{i \beta} \, .$$ The consciencious reader will now check that the two metrics \ref{met1} and \ref{met2} are the same.

### Embedding in Euclidean space

We have just seen that it is possible to embed isometrically the full hyperbolic plane into Minkowski space. Is it possible to do the same in Euclidean space? By this, I mean to find a (regular) map $f : \mathbb{H} \longrightarrow \mathbb{R}^3$ such that the metric induced on the image of $f$ by the Euclidean metric $\mathrm{d} s^2_{\textrm{Eucl}} = \mathrm{d} X_0^2 + \mathrm{d} X_1^2 + \mathrm{d} X_2^2$ be, possibly up to a change of coordinates, the metric $\mathrm{d} s^2_{\mathbb{H}}$ considered in the previous paragraphs. We will see in a minute that the answer is no. In fact, the hyperbolic plane is simply "too big" for $\mathbb{R}^3$. However, if we pick a constant $c>0$ and remove the part of $\mathbb{H}$ that satisfies $y < c$, then a solution is available. Because of a problem of universal covering, we also have to take a part of $\mathbb{H}$ of width $2 \pi c$ in the $x$ direction, for instance $0 \leq x < 2 \pi c$. Then, define $t=\mathrm{ArcCosh} \frac{y}{c}$ and \begin{eqnarray*} X_0 &=& t - \tanh t \\ X_1 &=& \mathrm{sech} \, t \cos \frac{x}{c}\\ X_2 &=& \mathrm{sech} \, t \sin \frac{x}{c} \end{eqnarray*} A cumbersome computation shows that the metric induced on that surface is indeed $\mathrm{d} s^2_{\mathbb{H}}$. This surface is called a (half) pseudosphere. Here is a picture (taken from wikipedia) Note that if you don't want to make the computation yourself, you can put this code in Mathematica:
t = ArcCosh[y/c];
x0 = t - Tanh[t];
x1 = Sech[t]*Cos[x/c];
x2 = Sech[t]*Sin[x/c];
Dt[x0]^2 + Dt[x1]^2 + Dt[x2]^2 /. Dt[c] -> 0 // PowerExpand // Simplify

and you should recover the metric of the hyperbolic plane.

So, as we see, it is possible to embed part of the hyperbolic plane in Euclidean $\mathbb{R}^3$. But if you want to embed all of it, this is simply not possible, by a theorem of Hilbert that says that there exists no complete regular surface of constante negative (Gaussian) curvature immersed in $\mathbb{R}^3$. However, for the curvature to be defined, you need to be able to differentiate at least twice on the manifold - this is hidden in the word "regular". To be more precise, a theorem of Efimov tells you that there is no $C^2$ embedding of a complete surface in $\mathbb{R}^3$ with curvature satisfying $K<-\epsilon$ for some strictly positive constant $\epsilon$. But if you're satisfied with a $C^1$ embedding, then the Nash-Kuiper theorem tells you that a $C^1$ isometric embedding of the full hyperbolic plane in Euclidean $\mathbb{R}^2$ does exist! And things become even more counter-intuitive when you know that this embedding can be put inside a ball of arbitrarily small size...

PS : I used this post to answer a question here.