Second derivative of $\det\sqrt{F^TF}$ with respect to $F$ I need to evaluate
$$
\frac{\partial^2W}{\partial F^2}(I)(F,F)=\sum_{i,j,k,l=1}^3\frac{\partial^2W}{\partial F_{ij}F_{kl}}(I)F_{ij}F_{kl}
$$
for $W=\det\sqrt{F^TF}$. Here $F\in\mathbb{R}^{3\times3}$.
From First derivative we know that
$$
\frac{\partial W}{\partial F}=WF^{-T}.
$$
To evaluate the sum above, I write
$$
\frac{\partial^2W}{\partial F^2}=W\left[\left(F^{-T}\right)^2+\frac{\partial F^{-T}}{\partial F}\right]․
$$
Here I got stack.
 A: Denote the gradient of $W$ as
$$\eqalign{
G &= \frac{\partial W}{\partial F}=F^{-T}W \cr
\frac{1}{W}\,G &= F^{-T} \cr
}$$
and find its differential and gradient
$$\eqalign{
dG &= F^{-T}\,dW + W\,dF^{-T} \cr
   &= F^{-T}(G:dF) - WF^{-T}\,dF^{T}\,F^{-T} \cr
   &= F^{-T}(G:dF) - WF^{-T}\,{\mathbb E}\,F^{-1}:dF^{T} \cr
   &= \frac{1}{W}\,\Big(G\star G - (G\,{\mathbb E}\,G^T):{\mathbb B}\Big):dF \cr\cr
\frac{\partial G}{\partial F}   &= \frac{1}{W}\,\Big(G\star G - (G\,{\mathbb E}\,G^T):{\mathbb B}\Big) \cr
}$$
where $(\star)$ denotes the dyadic product, $(:)$ denotes the Frobenius product, and $({\mathbb E, B})$ are fourth-order isotropic tensors with components 
$$\eqalign{
{\mathbb E}_{ijkl} &= \delta_{ik}\,\delta_{jl} \cr
{\mathbb B}_{ijkl} &= \delta_{il}\,\delta_{jk} \cr\cr
}$$
So, as expected, the second derivative of $W$ is a fourth-order tensor
$$\eqalign{
\frac{\partial^2 W}{\partial F^2} &= \frac{\partial G}{\partial F} \cr\cr
}$$
For the sake of completeness, I'll mention that there is one more isotropic tensor ${\mathbb A}=I\star I$ with components
$$\eqalign{
{\mathbb A}_{ijkl} &= \delta_{ij}\,\delta_{kl} \cr
}$$
We didn't need it for this problem, but it does illustrate how the dyadic product works.
Anyway, the isotropic tensors have properties which are useful for working with matrix expressions
$$\eqalign{
P:{\mathbb E} &= {\mathbb E}:P = P \cr
P:{\mathbb B} &= {\mathbb B}:P = P^T \cr
P:{\mathbb A} &= {\mathbb A}:P = I(I:P) = P\,\operatorname{tr}(P) \cr
P\cdot Q\cdot R &= (P\cdot {\mathbb E}\cdot R^T):Q \cr
P\cdot Q^T\cdot R &= (P\cdot {\mathbb E}\cdot R^T):{\mathbb B}:Q \cr
\cr
}$$
If you are uncomfortable with tensors, you can apply vectorization
$$\eqalign{
 g &= \operatorname{vec}(G) \cr
 f &= \operatorname{vec}(F) \cr
}$$ and then calculate $$\frac{\partial g}{\partial f}$$ which will be an ordinary matrix.
Update
Now I understand the whole question.  You wish to evaluate the above Hessian at $F=I$ and then fully contract it with the tensor $(F\star F)$. 
At $F=I$, the gradient reduces to $G=WI$ and Hessian reduces to
$$\eqalign{
{\mathbb H} &= \frac{\partial G}{\partial F} = W\,\Big(I\star I - {\mathbb E}:{\mathbb B}\Big) \cr 
 &= W\,({\mathbb A}-{\mathbb B}) \cr\cr
}$$
Now it can be contracted to a scalar value
$$\eqalign{
F:{\mathbb H}:F &= W\,\Big(F:({\mathbb A}-{\mathbb B}):F\Big) \cr
 &= W\,\Big(\operatorname{tr}(F)^2 - \operatorname{tr}(F^2)\Big) \cr
}$$
