Well the definition of the hyperbolic plane is not just a definition of a set. Indeed the hyperboloid model is diffeomorphic to a Euclidean plane. But not isometric!
So, (one of) the (right) definition is:
takes the component of the hyperboloid with $x>0$ and, as metric, restrict the Minkowski metric (that with signature ++-) (exercice: check that this is indeed positive definite)
As for theorems, item $iii)$ is now a theorem.
I suggest to look the great book by Benedetti and Petronio "Lectures on hyperbolic geometry" http://www.zbmath.org/?q=an:00107544
It is smaller, and perhaps less complete than Ratcliffe, but is very well written and readable. You will find there all the answer to your doubts.
Here you need to think of the pair (surface, metric) as the geometric object, not only of the surface. The upshot here is that the pairs (hyperboloid, euclidean metric) and (hyperboloid, Minkowski metric) are distinct geometric objects. The way to measure lengths and areas is not the same in these two "worlds". The first one has non-constant positive curvature, and the second one has constant negative curvature $-1$.
The reason is that the hyperboloid plays the role of a sphere with respect to the Minkowski metric, since it can be written as the solution set of $\langle p,p\rangle_L=-1$. Compare that with the sphere equation $\langle p,p \rangle_E=1$. If by now you're wondering whether the one-sheeted hyperboloid defined by $\langle p,p\rangle_L=1$ has something special about it, I'll tell you it does: it has constant curvature $1$ when equipped with the Minkowski metric (the one who correctly expresses it as a sphere).
In fact, here's an exercise for you to understand what happens: define $$g((x_1,y_1,z_1),(x_2,y_2,z_2))= 3x_1x_2+5y_1y_2+2z_1z_2$$and show that $M=\{(x,y,z)\in \Bbb R^3\mid 3x^2+5y^2+2z^2=1\}$ has constant curvature when equipped with $g$.
Solution for suggested exercise. Note that $M = \{ p \in \Bbb R^3\mid g(p,p)=1\}$. If $F(p) = g(p,p)$, then $M = F^{-1}(1)$, meaning that the $g$-gradient of $F$ is always $g$-normal to $M$. Since $${\rm d}F_p(v) = 2g(p.v) = g(2p,v),$$we get ${\rm grad}_gF(p)=2p$, and so $N(p)=p$ is an $g$-unit $g$-normal vector along $M$, just like for spheres in Euclidean space. Indeed, $M$ is a $g$-sphere. Note that the curvature of the Levi-Civita connection of $g$ in $\Bbb R^3$ is zero (indeed, all the Christoffel symbols w.r.t. the usual rectangular coordinates are zero -- this actually says that the Levi-Civita connection is the usual one). So, Gauss' Formula says that $$K(v,w) = \frac{g(I\hspace{-.1cm}I(v,v))g(I\hspace{-.1cm}I(w,w)) - g(I\hspace{-.1cm}I(v,w),I\hspace{-.1cm}I(v,w))}{g(v,v)g(v,w)-g(v,w)^2}.$$Just in the same way we do it for the sphere, you can check (or see e.g. page 101 in Chapter 4 in O'Neill's Semi-Riemannian Geometry with Applications to Relativity) that $I\hspace{-.1cm}I_p(v,w) = -g(v,w)N(p)$. Taking $(v,w)$ to be an orthonormal basis for the tangent plane $T_pM$, plugging in above gives $K(p)=1$ as expected.
Best Answer
The formula that I'm getting for $\Pi: H^{n}\rightarrow B^{n}$ is $$ \Pi(x_{1}, \cdots, x_{n+1}) = \frac{(x_{2}, \ldots, x_{n+1})}{1 + x_{1}} $$ This gives you $$ y_{i} = \frac{x_{i+1}}{1 + x_{1}} \qquad\text{ for }\qquad 1\le i\le n. $$
So yes, the idea of your formula is correct, but there are some minor problems with the indices.
As a side-note, there are variations based on how the Minkowski product is defined. If you had $\langle x, y \rangle = x_{1}y_{1} + \cdots + x_{n}y_{n} -x_{n+1}y_{n+1}$ where you treated the $x_{n+1}$-coordinate as the special one, you would have instead gotten the formula $$ \Pi(x_{1}, \cdots, x_{n+1}) = \frac{(x_{1}, \ldots, x_{n})}{1 + x_{n+1}}, $$ which might appear nicer because the indices are aligned better. John Lee's Riemannian Manifolds: An Introduction to Curvature book for example uses the latter convention.