We define two dimensional hyperbolic space via $\mathbb{H}^2=(H^+,(dx^2+dy^2)/y^2)$ with $$H^+=\{z\in\mathbb{C}|Im(z)\ge 0\}$$ and the action of $SL(2,\mathbb{R})$ via $$(\begin{pmatrix} a & b \\ c & d \end{pmatrix},z)\mapsto \frac{az+b}{cz+d}$$ It is now an exercise to show that $\operatorname{Iso}(\mathbb{H}^2)\subseteq SL(2,\mathbb{R})$, meaning that every Isometry of Hyperbolic two-space is contained in $SL(2,\mathbb{R})$.
The proof goes by showing that $SL(2,\mathbb{R})$ acts transititive on $\mathbb{H}^2$ and that the stabilizer of $i$ is $SO(2)$. This is used to calculate the derivative of the action of $SO(2)\le SL(2,\mathbb{R})$ via the above action on $H^+$. This gives $$D_if_A(v)=e^{-2\varphi i}v\text{ for } A=\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi\end{pmatrix}$$ using that $D_zf_A=\frac{1}{(cz+d)^2}$ for some $A\in SL(2,\mathbb{R})$.
Here in the solution of this exercise it is concluded, that $SO(2)$ acts transitvely on $T_iH^+$ and thus using the transitive action of $SL(2,\mathbb{R})$ on $H^+$ we have a transitive action on the whole tangent bundle $TH^+$. I really don't get how one is able to conclude this.
The rest of the proof than once uses transitivity of $SL(2,\mathbb{R})$ on $H^+$ to construct $f_T\circ f(i)=i$ for some isometry $f$ and than uses transititity of $SO(2)$ on $TH^+$ to construct $D_if_s(Df_T\circ f(1))=1$. This is than used together with geodesical completeness to conclude that $f_S\circ f_T\circ f=Id_2$ and thus that $f$ is fractional linear.
Edit:
Corrected typos:
- Missing $-$ sign
- Changed absolute value to $()$
To prove that $SO(2)$ acts transitively on $T_i H^+$, using the formula $$D_if_A(v)=e^{2\varphi i}v\text{ for } A=\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi\end{pmatrix} $$ all you have to do is prove that for each unit vector $w$ the equation $D_i f_A(v)=w$ can be solved for $v$. Since $D_i f_A$ is an orthnormal matrix, this should be straightforward: simply left multiply by the inverse of that matrix.
By the way, your formula $$D_z f_A=\frac{1}{|cz+d|^2}$$ is odd. You can think of $D_z f_A$ either as an orthogonal transformation of $\mathbb{R}^2$ or as a complex number, but either way it is not a real number in general.