It would probably take a whole book to fully cover the field of calculus of variations, functional/variational derivatives, Frechet derivatives, Euler-Lagrange equations, principle of stationary action, etc. We assume that OP has already perused several textbooks and websites on the topic, so a hastily composed Math.SE answer by us would likely not be able to do a better job than that.
In this answers we will instead merely try to present a compass of what is up and down in calculus of variations in order to navigate physical applications.
The main point is to understand which variables are independent and which variable are dependent in the variation process.
Let us for simplicity limit ourselves to point mechanics of a single particle in $\mathbb{R}^3$ (and leave generalizations to many particles, to non-trivial topology, to field theory, to higher derivatives, etc, to the reader).
I) On one hand, there is the Lagrangian function $L: \mathbb{R}^3\times \mathbb{R}^3\times [t_i,t_f]\to \mathbb{R}$. The Lagrangian function $({\bf r},{\bf v},t) \mapsto L({\bf r},{\bf v},t)$ is assumed to depend smoothly on the position ${\bf r}$, the velocity ${\bf v}$ and time $t$. Note in particular that the 3+3+1 arguments in the function $L$ are independent variables, and it makes sense to form partial derivatives
$$\tag{1} L_{\rm r}({\bf r},{\bf v},t)~:=~\frac{\partial L({\bf r},{\bf v},t)}{\partial {\bf r}}, \quad L_{\rm v}({\bf r},{\bf v},t)~:=~\frac{\partial L({\bf r},{\bf v},t)}{\partial {\bf v}}, \quad L_{t}({\bf r},{\bf v},t)~:=~\frac{\partial L({\bf r},{\bf v},t)}{\partial t}. $$
II) On the other hand, there is the action functional $S: C^1([t_i,t_f]; \mathbb{R}^3) \to \mathbb{R}$, defined as
$$\tag{2} S[\gamma]~:=~ \int_{t_i}^{t_f}\!dt~ L(\gamma(t),\dot{\gamma}(t) ,t), $$
where $\gamma:[t_i,t_f]\to \mathbb{R}^3$ is the path of the particle in position space. Here a dot denotes time differentiation.
Note that the velocity profile $\dot{\gamma}$ is not independent of the position profile $\gamma$.
For a well-posed variational problem (i.e. with appropriate boundary conditions for the path $\gamma$ at initial and final times), one may form the functional derivative
$$\tag{3} \frac{\delta S[\gamma]}{\delta \gamma(t)}$$
of the action functional $S[\gamma]$ wrt. to the curve position $\gamma(t)$ at time $t$. Often the functional derivative (3) can be identified with the Euler-Lagrange expression
$$\tag{4} \frac{\delta S[\gamma]}{\delta \gamma(t)}
~=~ L_{\bf r}(\gamma(t),\dot{\gamma}(t),t)
-\frac{dL_{\bf v}(\gamma(t),\dot{\gamma}(t),t)}{dt} .$$
For more information, see e.g. this Phys.SE post.
III) Concerning OP's last expression, a functional derivative
$$\tag{5} \frac{\delta S[\gamma]}{\delta \dot{\gamma}(t)}\qquad\qquad \longleftarrow\text{(Wrong!)}$$
of the action $S[\gamma]$ wrt. the velocity $\dot{\gamma}(t)$ does not readily make sense$^1$ in point mechanics, if at all, because we can not vary $\dot{\gamma}$ independently of $\gamma$.
IV) Note that it is possible to define a functional derivative
$$\tag{6}\frac{\delta L(\gamma(t),\dot{\gamma}(t) ,t)}{\delta \gamma(t^{\prime})}
$$
of the Lagrangian function in a distributional sense. Here $t$ and $t^{\prime}$ are two independent times.
V) However, if we try to apply the definition of a functional derivative
$$\tag{7} \frac{\delta L(\gamma(t),\dot{\gamma}(t) ,t)}{\delta \gamma(t)},\qquad\qquad\longleftarrow\text{(Wrong!)},
$$
at the same time $t$, we would get a meaningless result. Nevertheless, be aware that many authors confusingly use the 'same-time' functional derivative (7) as a shorthand notation for the Euler-Lagrange expression (4), or the functional derivative (3), cf. e.g. my Phys.SE answers here and here.
--
$^1$ Note however, that in field theory (as opposed to point mechanics) that a functional derivative
$$\tag{8} p({\bf r},t)~:=~ \frac{\delta L(t)}{\delta v({\bf r},t)} $$
of the Lagrangian function
$$\tag{9} L(t)~=~\int_{\mathbb{R}^3} \!d^3r~{\cal L}({\bf r},t) $$
wrt. the velocity field $v({\bf r},t)$ at the same time $t$ does make sense in field theory. (Some of the arguments in eqs. (8) and (9) are not written explicitly.) Actually, this field theoretic subtlety (8) may have been OP's main question all along. This is explained e.g. in this and this Phys.SE posts.]
Best Answer
I see no other interpretation than $G$ being a function-valued functional, e.g. $G:C^\infty(\mathbb R)\to C^\infty(\mathbb R).$ An example of such a functional is $G[\phi](x)=e^{-x^2/2}\phi'(x)^2.$
Then $F[G[\phi]]$ makes sense and $$ \frac{\delta F[G[\phi]]}{\delta\phi(x)} = \int \frac{\delta F[G[\phi]]}{\delta G[\phi](y)} \frac{\delta G[\phi](y)}{\delta\phi(x)} \, dy . $$