Let $$u_{tt}-u_{xx}= u-u^3 ,\: (t,x)\in \mathbb{R}\times \mathbb{R}.$$
I know that this equation admits wave traveling solution of the form $u(x,t)=\varphi(x-ct)$, where $ c\in \mathbb{R}$ is a constant and $\varphi$ is $L>0$ periodic. Moreover, using this question I found the operator linearized around $\varphi$ given by $$\mathcal{L}=(1-c^2)\frac{d^2}{dx^2}+(1-\varphi^2).$$
However, I want to find $ \mathcal{L} $ using conserved quantities $E$ and $F$, that is, write $$G''(\varphi):=E''(\varphi) -cF''(\varphi)=\mathcal{L}.$$ But I was unable to find such conserved quantities. A candidate for conservede quantitie $F$ I think is $$F(u)=\frac{1}{2}\int_{0}^{L} u^2\;dx.$$
This is more of a comment than an actual answer, but: the idea is to use that derivatives by themselves are linear operators and to take the total derivative of the differential equation with the variable $u$. So the linearization of your ODE centered at $u$ is $$v_{tt} - v_{xx} = v- 3u^2v,$$seen as an equation for the "tangent vector" $v$ at $u$. I do not know how you got a single-variable differential operator for the linearization (nor how one would find a linearization via "conserved quantities", this seems like overcomplicating it). Using your notation, the linearized equation written in the form $\mathcal{L}v=0$ should be $$\mathcal{L} = (\partial_t)^2 - (\partial_x)^2 + (3u^2-1).$$