general relativity - Is there a natural (suitable) definition for functional derivative in Curved space time

If

$$\delta S = \int \sqrt g F[\phi] \delta \phi\tag{1}$$

Then is it natural to define the functional derivative as follows,

$$\frac{\delta S}{\delta \phi} = F[\phi].\tag{2}$$

In particular does this definition satisfy the commutativity of the functional derivatives.

I understand that this is not the standard definition of a functional derivative, but if I define this way this makes a certain calculation I am doing much easier to control. So to Clarify what I want to know is that if I define the 'functional derivative' this way then if

$$S = \int \sqrt{g} L[\phi, g_{\mu \nu}]\tag{3}$$ then

is the following true?

$$\frac{\delta}{\delta\phi} \frac{\delta}{\delta g_{\mu \nu}} S = \frac{\delta}{\delta g_{\mu \nu}} \frac{\delta}{\delta\phi} S \tag{4}$$

This is my procedure for computing the second functional derivative suppose

$$\frac{\delta}{\delta \phi}S = E[\phi, g_{\mu \nu}]\tag{5}$$

and

$$\frac{\delta}{\delta g_{\mu \nu}}S = E^{\mu \nu}[\phi, g_{\mu \nu}]\tag{6}$$

(equations of motion) Then to compute second functional derivative we write, eg.

$$E[\phi, g_{\mu \nu}](x) = \int \sqrt{g} d^4 y E(y) \hat \delta(x-y)\tag{7}$$

where

$$\hat \delta (x-y) = \frac{\delta(x-y)}{\sqrt{g}}\tag{8}$$ is the generalised delta function. After this we can use the same definition to compute

$$\frac{\delta} {\delta g_{\mu \nu}} \int \sqrt{g} d^4 y E(y) \hat \delta(x-y)\tag{9} .$$ Of course now the result would involve delta functions.