Question on Weil-Petersson metric on Teichmuller space

I'm reading Ahlfors' original articles about Weil-Petersson metric: "Some remarks on Teichmüller's space of Riemann surfaces" and "Curvature properties of Teichmüller's space".

The tangent space at a point $C$ (here identified with a Riemann surface of genus $g$) of the Teichmuller space can be identified with the space of Beltrami differentials on $C$ and the author defines the Weil-Petersson hermitian form as $\langle \nu,\mu\rangle=\int_C\nu\varphi(\mu)dxdy$, for every Beltrami differentials $\nu,\mu$ on $C$, where $\varphi(\mu)$ is the quadratic differential corresponding to $\mu$.

On the Teichmuller space $\mathcal{T}_g$ there are complex coordinates $t_1,\dots,t_n$, because every Beltrami differential $\mu$ can be written $\mu=t_1\nu_1+\dots+t_n\nu_n$, where $\nu_1,\dots,\nu_n$ is a base. Then author then wants to write the Weil-Petersson form in this coordinates as $\sum_{\alpha,\beta}g_{\alpha\overline{\beta}}(\mu)dt_\alpha d\overline{t}_\beta$.

What I don't get is why he writes $g_{\alpha\overline{\beta}}(\mu)=\langle L^\mu\nu_\alpha,L^\mu\nu_\beta\rangle$

The definition of $L^\mu\nu$ is the following. Consider $f^\mu:\hat{\mathbb{C}}\rightarrow\hat{\mathbb{C}}$ such that $\partial_{\overline{z}}f^\mu=\mu\partial_zf^\mu$, then, given three Beltrami differentials $\mu,\rho$ and $\lambda$, we write $\rho=\mu|\lambda$ iff $f^\mu=f^\rho\circ f^\lambda$ which is true iff $\rho\circ f^\lambda=\frac{\mu-\lambda}{1-\overline{\lambda}\mu}(\partial_zf^\lambda/|\partial_zf^\lambda|)^2$.

Then, by abuse of notation, we can define the function on Beltrami differentials $\rho:\mu\mapsto \mu|\lambda$ and $\frac{\partial}{\partial \nu}\rho(\lambda):=lim_{t\rightarrow 0}\frac{\rho(\lambda+t\nu)-\rho(\lambda)}{t}=L^\lambda\nu=(\nu\frac{(\partial_zf^\lambda)^2}{|\partial_zf^\lambda|^2-|\partial_\overline{z}f^\lambda|^2})\circ(f^\lambda)^{-1}$

Why this definition of $L^\mu\nu$ should imply that, imposing $g_{\alpha\overline{\beta}}(\mu)=\langle L^\mu\nu_\alpha,L^\mu\nu_\beta\rangle$, we get the metric $\int_C\nu\varphi(\mu)dxdy$?

One way to define Teichmueller space is to fix a Riemann surface $X$, with a fixed complex structure, and define the space of all Beltrami differentials $\mathcal{M}(X)$ on $X$. Notice that in this way we have fixed a base point $X$ of Teichmueller space with a fixed "background" complex structure. By a Beltrami differential I mean a $(-1,1)$ form $\mu(z)\frac{d\bar{z}}{d z}$ such that $\|\mu\|_{\infty} < 1$. By an infinitesimal Beltrami differential I simply mean a $(-1,1)$ form without the latter inequality, all of which form the set $\mathcal{IB}(X)$. Then one way to define Teichmueller space is $\mathcal{T}(X) = \mathcal{M}(X) / \sim$, where two Beltramies $\mu_1$ and $\mu_2 \in \mathcal{M}(X)$ are equivalent, i.e. $\mu_1 \sim \mu_2$, exactly when there is a quasi-conformal homeomorphism $f : X \to X$ homotopic to identity such that $$\mu_1 = f^*\mu_2 = \frac{\bar{\partial} f + (\mu_2\circ f) \overline{\partial f}}{{\partial} f + (\mu_2 \circ f) \partial\bar{f}}.$$ Furthermore, the tangent space $T_X\mathcal{T}(X)$ of $\mathcal{T}(X)$ at the base point $X$ can be thought of as $\mathcal{IB}(X)/\sim$.
Now, if by $X^{\mu}$ I denote the Riemann surface with complex structure induced by the Beltrami $\mu$ on $X$, then the tangent space of the Teichmueller sapce at that point is $T_{X^{\mu}}\mathcal{T}(X) = T_{[\mu]}\mathcal{T}(X) = \mathcal{IB}(X^{\mu})/\sim$. The Wiel-Petersson hermitian form, calculated at the point $[\mu]$, is defined as $$\langle\nu_1, \nu_2\rangle_{\mu} = \int_{X^{\mu}} \nu_1 \varphi(\nu_2),$$ where $\nu_1$ and $\nu_2 \in \mathcal{IB}(X^{\mu})$ are infinitesimal Beltramies on $X^{\nu}$ (not on $X$!). This is the reason for which one needs to use the formulas you are asking about. One needs to look at all the infinitesimal Beltramies $\nu = \sum_{\alpha} t_{\alpha}\nu_{\alpha}$ on the base point Riemann surface $X$ from the perspective of the Riemann surface determined by $[\mu] \in \mathcal{T}(X)$, i.e. to look at the infinitesimal Beltriemies $\nu|_{\mu} = \sum_{\alpha} t_{\alpha}\nu_{\alpha}|_{\mu}$ on the surface $X^{\mu},$ i.e. $\nu|_{\mu} \in \mathcal{IB}(X^{\mu})$. In order to do that, first one needs a way to move all Beltramies from $\mathcal{M}(X)$ to $\mathcal{M}(X^{\mu})$ by establishing an appropriate isomorphism. Then, by differentiating that isomorphism at the point $[\mu]$ one would find out how infinitesimal Beltramies are moved from $X$ to $X^{\mu}$ isomorphically, i.e. one would obtain the tangent isomorphism. Consequently, one would have an identification between $\mathcal{IB}(X^{\mu})/\sim$ and $\mathcal{IB}(X)/\sim$.
Given a Beltrami $\lambda$ denote by $f^{\lambda} : X \to X^{\lambda}$ the quasi-conformal map (unique up to post-composition with a holomorphic isomorphism) that determines the Beltrami $\lambda = \frac{\bar{\partial} f^{\lambda}}{\partial f^{\lambda}}$ on $X$. Also denote by $f^{\lambda|_{\mu}} : X^{\mu} \to X^{\lambda}$ the quasi-conformal map that determines the (intermidiate) Beltrami $\lambda|_{\mu} = \frac{\bar{\partial} f^{\lambda|_{\mu}}}{\partial f^{\lambda|_{\mu}}}$ on $X^{\mu}$ (not on $X$!). Then, (technically up to holomorphic isomorphism) $f^{\lambda} = f^{\lambda|_{\mu}} \circ f^{\mu} : X \to X^{\lambda},$ where $f^{\lambda|_{\mu}} : X^{\mu} \to X^{\lambda}.$ After differentiating the latter identity between the three quasi-conformal maps, $df^{\lambda} = df^{\lambda|_{\mu}} \circ df^{\mu}$, and then combining it with the expressions of the type $df(z) = \partial f(z) dz + \bar{\partial} f(z) d\bar{z}$, one obtains again the formula, already featured above,
$$\lambda = f^{\mu*}(\lambda|_{\mu}) = \frac{\bar{\partial} f^{\mu} + (\lambda|_{\mu} \circ f^{\mu}) \overline{\partial f^{\mu}}}{{\partial} f^{\mu} + (\lambda|_{\mu} \circ f^{\mu}) \partial\overline{f^{\mu}}}.$$ By solving for $\lambda|_{\mu} \circ f^{\mu}$ and using the fact that $\bar{\partial} f^{\mu} = \mu \partial f^{\mu}$, one obtains exactly the expression $$\lambda|_{\mu} \circ f^{\mu} = \frac{\lambda - \mu} {1 - \bar{\mu}\lambda}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2.$$ This is basically the map from $\mathcal{M}(X)$ to $\mathcal{M}(X^{\mu})$ sending Beltramies on $X$ to Beltramies on $X^{\mu}$. To obtain a tangent vector at $[\mu] \in \mathcal{T}(X)$, take an infinitesimal Beltrami $\nu$ on $X$ and form the one parameter family $\mu_t = \mu + t \nu$. On $X^{\mu}$ this family becomes $$\mu_t|_{\mu} \circ f^{\mu} = \frac{\mu + t \nu - \mu} {1 - \bar{\mu}(\mu + t \nu)}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2 = t \frac{\nu } {1 - \bar{\mu}(\mu + t \nu)}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2.$$ To obtain the tangent vector $\nu|_{\mu}$ at the point $\mu$ simply differentiate the last identity with respect to $t$ and then set $t=0$. The result is $$\nu|_{\mu} \circ f^{\mu} = \nu \frac{(\partial f^{\mu})^2} {|\partial f^{\mu}|^2 - |\bar{\partial} f^{\mu}|^2}.$$ After pre-composing with $(f^{\mu})^{-1}$, one obtains exactly the formula you mention $$L^{\mu}\nu = \nu|_{\mu} = \left(\nu \frac{(\partial f^{\mu})^2} {|\partial f^{\mu}|^2 - |\bar{\partial} f^{\mu}|^2}\right)\circ (f^{\mu})^{-1}.$$ That is why the Weil-Petersson form should look like $$\langle\nu_{\alpha}|_{\mu}, \nu_{\beta}|_{\mu}\rangle_{\mu} = \langle L^{\mu}\nu_{\alpha}, L^{\mu}\nu_{\beta}\rangle_{\mu} = \int_{X^{\mu}} \nu_{\alpha}|_{\mu} \varphi(\nu_{\beta}|_{\mu}) = \int_{X^{\mu}} L^{\mu} \nu_{\alpha} \varphi(L^{\mu} \nu_{\beta}),$$ for any $\nu_{\alpha}$ and $\nu_{\beta}$ from $\mathcal{IB}(X)$ (notice that the base point is $X$!).