The action of Mod$(T)$ on Teich($T$)

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This is Proposition 7.1.5 in Martelli's book which I am getting confused.

The action of MCG(T) on $\tau(T)$ is the following action of $SL_2(\mathbb{Z})$ on $\mathbb{H}^2$ as Mobius transformations: $$\varphi=\begin{bmatrix}a&b\\c&d\end{bmatrix}:z\mapsto \frac{az-b}{-cz+d}.$$

Let me clarify my notations: let $S_g$ be a closed orientable surface of genus $g$, and $\tau(S_g)=\{(X,f)\}/\sim$ be the Teichmuller space where $f:S_g\to X$ is the marked homeomorphism. When $g=1$, we consider the set of flat metrics on $S_1$ and hyperbolic when $g\geq 2$. MCG($S_g$) denotes the mapping class groups. The action of $\varphi:S_g\to S_g$ in MCG$(S_g)$ on $\tau(T)$ is defined as $(X,f)\mapsto(X,f\circ\varphi^{-1})$.

$\textbf{Assisted by my professor and thanks to all the comments}$, I have the following result now. Please let me know if I missed anything.

The key I am missing is really the what is going on (isomorphism) between $\tau(S)$ and $\mathbb{H}^2$. This proposition is in fact about the equivariance i.e. $\rho(\varphi\cdot(X,f))=\varphi\cdot(\rho(X,f))$ where $\rho:\tau(S)\to\mathbb{H}^2$.

Now, let us consider the flat torus $T^2$. Given an element $(X,f)\in\tau(T^2)$ and let $\rho(X,f)=z$. Assume without loss of generality that $(0,1),(1,0)$ are generators of $\pi_1(T^2)$. Then $f$ sends $(1,0)$ to $1$ and $(0,1)$ to $z$ up to rotation and rescaling. Now, $\varphi(X,f)=(X,f\circ\varphi^{-1})$ where $f\varphi^{-1}$ sends $(1,0),(0,1)$ to $f(d,-c),f(-b,a)$ respectively. Then $f(d,-c)=f(d(1,0)-c(0,1))=d-cz$ and similarly $f(-b,a)=-b+az$. Hence, we have $\varphi(z)=\frac{-b+az}{d-cz}$ as desired.