I need to find a relation between the tangent stiffness $L_1$ of the first Piola-Kirchhoff stress tensor with the tangents stiffness $L_2$ of the second Piola-kirchoff stress tensor. They are defined as
\begin{align} &L_1 = \frac{dP}{dF}\, ,\\ &L_2 = 2\frac{dP}{dC}\, , \end{align}
with $F$ being the deformation gradient and $C$ being the right Cauchy-Green tensor.
If I got $L_1$, there is a way to compute $L_2$?
Ignoring subtleties regarding differentiation in two separate tangent spaces, and also avoiding complications arising from non-Euclidean coordinates, we can proceed as follows.
The derivative of $\boldsymbol{P}$ with respect to $\boldsymbol{C}$ is $$ \frac{\partial\boldsymbol{P}}{\partial\boldsymbol{C}} = \frac{\partial\boldsymbol{P}}{\partial\boldsymbol{F}} : \frac{\partial\boldsymbol{F}}{\partial\boldsymbol{C}} $$ where, for two fourth-order tensors $\mathbb{A}$ and $\mathbb{B}$, the contraction operaion is defined as (summation implied on repeated indices) $$ \mathbb{A} : \mathbb{B} = A_{ijk\ell} B_{k\ell mn} $$ We need to find $\partial\boldsymbol{F}/\partial\boldsymbol{C}$. Let us start with the definition of $\boldsymbol{C}$: $$ \boldsymbol{C} = \tfrac{1}{2} (\boldsymbol{F}^T \cdot \boldsymbol{F} - \boldsymbol{I}) $$ where the partial inner product of two second order tensors $\boldsymbol{A}$ and $\boldsymbol{B}$ is defined as $$ \boldsymbol{A}\cdot\boldsymbol{B} = A_{ij} B_{jk} $$ Then, $$ \frac{\partial\boldsymbol{C}}{\partial\boldsymbol{F}} = \tfrac{1}{2}\left[ \frac{\partial\boldsymbol{F}^T}{\partial\boldsymbol{F}}\cdot\boldsymbol{F} + \boldsymbol{F}^T \cdot\frac{\partial\boldsymbol{F}}{\partial\boldsymbol{F}}\right] $$ where the contraction between a fourth order tensor $\mathbb{A}$ and a second order tensor $\boldsymbol{B}$ is defined as $$ \mathbb{A} \cdot \boldsymbol{B} = A_{ijkm} B_{m\ell} \quad \text{and} \quad \boldsymbol{B} \cdot \mathbb{A} = B_{im} A_{mjk\ell} $$ Expressions for derivatives can be found at https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Derivative_of_a_second-order_tensor_with_respect_to_itself. Using these, we have $$ \frac{\partial\boldsymbol{F}^T}{\partial\boldsymbol{F}}\cdot\boldsymbol{F} = \delta_{im}\delta_{jk} F_{m\ell} = F_{i\ell} \delta_{jk} $$ and $$ \boldsymbol{F}^T \cdot\frac{\partial\boldsymbol{F}}{\partial\boldsymbol{F}} = F_{mi} \delta_{mk} \delta_{j\ell} = F_{ki} \delta_{j\ell} $$ which leads to $$ \left[\frac{\partial\boldsymbol{C}}{\partial\boldsymbol{F}}\right]_{ijk\ell} = \tfrac{1}{2}(F_{i\ell} \delta_{jk} + F_{ki} \delta_{j\ell}) $$ Finally, $$ \mathbb{L}_2 = 2\frac{\partial\boldsymbol{P}}{\partial\boldsymbol{F}} :\left[\frac{\partial\boldsymbol{C}}{\partial\boldsymbol{F}}\right]^{-1} $$ where the inverse of the fourth-order tensor can be computed using the approach given in https://math.stackexchange.com/questions/1624955/the-inverse-of-a-fourth-order-tensor.
It may actually be simpler to start with the definition of $\boldsymbol{P}$ and $\boldsymbol{S}$, e.g., $$ \boldsymbol{P} = \det(\boldsymbol{F}) \,\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma} ~,~~ \boldsymbol{S} = \det(\boldsymbol{F})~\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T} $$ and compute the derivative directly using $$ \frac{\partial}{\partial\boldsymbol{A}}\det(\boldsymbol{A}) = \det(\boldsymbol{A})~\left[\boldsymbol{A}^{-1}\right]^\textsf{T} $$ and $$ \frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1} \implies \frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = - A^{-1}_{ik}~A^{-1}_{lj} $$ That's left as an exercise for the reader :)
