The linearization of a system and the derivative of operator. Firstly, in red line 1 of the picture below,whether it means make a variable substitution  $\widetilde {g}_{ij}=f(x)g_{ij}$? Because in my opinion, the linearization of something is to make a linearization of a function.Example,the linearization of a function $f(x)$ at point $p$ is $f(x)\approx f(p)+\nabla f|_p\cdot(p-x)$.But the 1.2.1 seems be not a function, how to make linearization?
Secondly,I think the $E$ in the below picture is a operator. Why a operator has derivative? As a operator ,it hasn't any  variable ,how to define it's derivative ?
I'm a layman about this ,and really hope a detail answer or hint.Thanks very much.
The below picture is from the 178th page of this paper.

 A: An operator is a function, just between more exotic spaces than you might be used to dealing with. This $E$ is a function that sends $2$-tensor(-field)s to $2$-tensors, and the set of all $2$-tensors is a vector space, so it shouldn't be too strange to imagine we can take directional derivatives of $E$.
The notation $DE(g_{ij})\tilde g_{ij}$ means the derivative of $E$ at the point $g$ in the direction $\tilde g$ - so conceptually this is $\nabla E|_g \cdot \tilde g$. 
This $D$ is the generalised directional derivative defined by
$$ DE(p)q = \frac d {ds} \Big|_{s=0} E(p + sq). $$
Now how do we linearize our equation? If we wanted to linearize an equation $h(x) = k(x)$ of real-valued functions about a known solution $h(x_0)=k(x_0)$, we'd simply rearrange to $f(x) = h(x) - k(x) = 0$ and then apply the linear approximation to $f$ yielding $$f(x_0) + \nabla f|_{x_0} \cdot (x-x_0) = \nabla f|_{x_0} \cdot (x-x_0) = 0.$$
We can do exactly the same thing here - write the equation as $f(g) = \partial_t g - E(g) = 0$, and then linearize about a known solution $f(g) = 0$, yielding $$Df(g)\tilde g = \partial_t \tilde g - DE(g)\tilde g=0.$$
