Your questions 1 and 2 can be answered if you consider one more geometrical aspect of the group of the one-dimensional Lorentz bosts and that is the orbit of a single point. In other words, the set obtained by transforming one fixed point, say, (0, 1), by Lorentz boosts of all possible $\beta$'s. If you would apply several transformations successively, this is a trajectory the point would follow and never leave, so we have an important invariant.
You should be getting one arm of a hyperbola with asymptotes given by the boundaries of a light cone. The interval $\Delta s$, measured from the reference point in the origin, indexes the possible hyperbolas: it is their semi-major axis. This helps to understand why $s$ is defined the way it is, its geometrical meaning and its conservation under boosts.
The semi-axis of a hyperbola going through a given point is also one half of its so-called hyperbolic coordinates. The other component is the hyperbolic angle, which, unlike polar angle, can reach any value in $\mathbb R$. Any Lorentz boost is just a shift of the hyperbolic angle—it adds a constant $\kappa$ to the h.a. of any transformed point. This constant can be found in the boost matrix if you rewrite it like
$$\begin{pmatrix}
\cosh{\kappa} & 0 & 0 & -\sinh{\kappa} \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\sinh{\kappa} & 0 & 0 & \cosh{\kappa} \\
\end{pmatrix}.$$
(This is always possible as $\gamma^2 - (\gamma\beta)^2 \equiv 1$.) All these facts, together with the similarity of the above matrix with a rotational matrix, justify calling boosts "some sort of" rotations in the $(t, x)$ plane.
I'm afraid I will leave your third question unanswered as it is largely philosophical and out of my league. However, you could easily find that if you accept the idea of Lorentz transforms, the special relativistic scalar product is the only choice that fits perfectly into the picture. Most importantly, because it stays invariant with respect not only with boosts, but also all other space-time transforms considered in the SR.
Timaeus's answer could be correct. The $\Lambda$ matrix from your book may have been intended as a passive transformation (one that acts on the coordinate system) and you mistakenly used it as an active transformation (one that acts on the object). Alternatively, the $\Lambda$ matrix from your book may have been intended as the active transformation of a co-vector and my answer below addresses that.
Because boosts are not orthogonal matrices, you have to think about the different way co-vectors and contra-vectors transform. The $\Lambda$ matrix you wrote down is the +v boost for a co-vector. You then applied it to a vector and incorrectly interpreted the result as if it were a contra-vector. The co-vector $U_\alpha$ with +v velocity really does have $-v\gamma$ in it as you found. You must raise U's index with the metric to see what the contra-variant $U^\beta$ is, and you see the +v that you expected.
$$U^\beta = \eta^{\beta \alpha}U_\alpha$$
$$\begin{bmatrix}\gamma \\v\gamma \\0 \\0\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 &-1 & 0 \\ 0 & 0 & 0 &-1\end{bmatrix} \begin{bmatrix}\gamma \\-v\gamma \\ 0 \\ 0\end{bmatrix}$$
A contra-vector transforms as
$$Unew^\alpha = \Lambda^\alpha _{-\beta} Uold^\beta$$
Then a co-vector transforms like
$$Unew_\alpha = Uold_\beta(\Lambda^{-1})^\beta _{-\alpha}$$
This matrix boosts contra-vectors by +v. I have redefined your $\Lambda$ to be the matrix that gives a +v boost to a contra-vector.
$${\Lambda=\begin{bmatrix}\gamma & v\gamma & 0 & 0 \\
v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1
\end{bmatrix}}$$
$$\begin{bmatrix}\gamma \\v\gamma \\0 \\0\end{bmatrix}
=\begin{bmatrix}\gamma & v\gamma & 0 & 0 \\
v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
This matrix boosts co-vectors by +v.
$${(\Lambda^{-1})^T=\begin{bmatrix}\gamma & -v\gamma & 0 & 0 \\
-v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1
\end{bmatrix}}$$
$$\begin{bmatrix}\gamma \\-v\gamma \\0 \\0\end{bmatrix}
=\begin{bmatrix}\gamma & -v\gamma & 0 & 0 \\
-v\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
Notice that rotations R are orthogonal which means $(R^{-1})^T=R$ and the transformations for co-vectors and contra-vectors are the same. Hence we don't talk about co-vectors and contra-vectors when doing rotations.
Best Answer
Take units $c=1$.
You have $U_0^2-\vec U^2=1$, that is $\gamma^2(1-\beta^2)=1$. With some basic transformations, you will get : $\frac{\gamma - 1}{\beta^2}= \frac{\gamma^2}{\gamma + 1}$
Now, from your Wikipedia matrix, you have obvious term, $ U_0 =\gamma , U_i =\gamma \beta_i$
You have $(\gamma -1) \frac{\beta_i\beta_j}{\beta^2} = \frac{\gamma^2}{\gamma + 1}\beta_i\beta_j = \frac{U_i U_j}{U_0 + 1}$
Finally, $1 + (\gamma -1) \frac{\beta_i^2}{\beta^2} = 1 + \frac{U_i^2}{U_0 + 1}$