Given $z_0,z_1$ in the upper half plane, there is $g\in SO(2,\mathbb{R})$ such that $g(i)=z_0$ and $g'(i)=z_1$. Let $\mathbb{H}:=\{z\in\mathbb{C}:\Im z>0\}$ be the upper half-plane. We can consider the action of the group $SL(2,\mathbb{R})$ of real $2\times 2$ matrix with determinant equal to $1$ given by Möbius transformations:
$$\left(\begin{matrix}
a & b \\
c & d 
\end{matrix}\right)\cdot z :=\frac{az+b}{cz+d}$$
Its easy to see that this is well defined and constitutes an action. I'd like to see that, given $z_0,z_1\in\mathbb{H}$, there exists some $g\in SL(2,\mathbb{R})$ such that $g(i)=z_0$ and $g'(i)=z_1$. Since $g'(z)=\frac{1}{(cz+d)^2}$, this reduces to:
$$\frac{ai+b}{ci+d}=z_0=x_0+iy_0$$
$$\frac{1}{(ci+d)^2}=z_1=x_1+iy_1$$
The brute force approach of trying to solve this for $a,b,c,d$ (with the additional equation $ad-bc=1$ since $g\in SL(2,\mathbb{R})$) gives rise to an ugly system of non-linear equations, and my book claims this is all very clear, so I assume there's an easier way of doing it. 
 A: The Iwasawa decomposition for $\mathrm{SL}(2,\mathbb{R})$ more or less is the following: we have three subgroups called $K,A,N$ (standing for kompakt, abelian, nilpotent) which are $K=\mathrm{SO}(2)$, $A$ is comprised of the diagonal  matrices with positive entries, and $N$ is comprised of unitriangular matrices. Then the multiplication map $K\times A\times N\to\mathrm{SL}(2,\mathbb{R})$ given by $(k,a,n)\mapsto kan$ is a diffeomorphism of smooth manifolds (albeit not a group homomorphism).
Notice that $B=AN$ is the subgroup of upper triangular matrices, and acts by real affine transformations of the plane, as $(\begin{smallmatrix} a & b \\ 0 & a^{-1}\end{smallmatrix})z=a^2z+b$. The real scalings $z\mapsto a^2z$ are simply dilations of the plane, while the translations $z\mapsto z+b$ shift the plane left and right. Then $B$ acts regularly on the upper half plane: given any two $z,w\in\mathbb{H}$ there is a unique $g\in B$ with $gz=w$.
To see this, note that we can uniquely write $g$ as a composition of a scaling and then a translation (in that order, since $AN=NA$). The scaling is uniquely determined by the condition that it must force $z$'s imaginary part to match that of $w$, and then the translation from there to $w$ is also uniquely determined.
Next, notice $K=\mathrm{SO}(2)$ is the point-stabilizer of $i$, and its action by hyperbolic rotation acts by rotations on the unit tangent space $UT_i\mathbb{H}=S^1$ (albeit in a $2$-to-$1$ fashion, since $K$ contains $\pm I$, the kernel of $\mathrm{SL}(2,\mathbb{R})$'s action, the reason it descends to a $\mathrm{PSL}(2,\mathbb{R})$ action). This implies that $\mathrm{PSL}(2,\mathbb{R})$ acts regularly on the unit tangent bundle $UT\mathbb{H}$: first notice $KAN=NAK$, so that every $g\in\mathrm{PSL}(2,\mathbb{R})$ is uniquely a composition of a hyperbolic rotation around $i$, a real scaling away from $0$, and a real translation left/right, in that order. The latter two parts are uniquely determined by how to get from one point to the next, and the hyperbolic rotation is uniquely determined by how to get to the new unit tangent vector from the old.
