Take the category of all G-sets for different groups $G$. Each $G$-set is the groupoid $G \times \Omega \to \Omega$ (the first projection is the source and the action map is the target).
I am not sure how to understand this statement. One way to understand it is the following:
Denote
$\mathcal D:=\{p| G \times \Omega \xrightarrow[]{p} G\}$
$\mathcal C:=\{f| G \times \Omega \xrightarrow[]{f} \Omega\}$
From the statement, we deduce that $\mathcal D$ is a fibred in groupoids over $\mathcal C$.
If the above is correct, Can we clearly describe the functor $F:\mathcal D\to \mathcal C$?
My appologies if the question is not clear enough, it isn't either in my mind.
I would agree generally that the question is sort of unclear. In particular, $\mathcal{D}$ and $\mathcal{C}$ in your question are the same, except that you've renamed a variable, which doesn't really change what they are in any meaningful sense.
That said, I can explain the statement.
Let's start with
I assume $\Omega$ is a $G$-set here, otherwise this doesn't make sense.
Background on internal groupoids/categories
Recall that an internal category to $\mathbf{Set}$ is a set $M$ of morphisms, a set $O$ of objects, and three maps, $s,t : M\to O$ that specify the source and target of each morphism, and $i : O\to M$ that picks out the identity morphism for each object. There should also be a composition morphism $\mu : M\times_{s,O,t} M \to M$ (where the fiber product over $O$ is taken with respect to $s$ in the first variable and $t$ in the second) such that all of the axioms defining a category hold.
An internal groupoid has one more morphism, inversion, $\tau : M \to M$ that picks out the inverse of each morphism. This should also satisfy the appropriate axioms.
An interpretation of this statement
The parenthetical is saying that a $G$-set $\Omega$ defines an internal groupoid. The morphism set is $G\times \Omega$, the object set is $\Omega$, the source morphism $s:G\times \Omega\to \Omega$ is the projection onto $\Omega$ (which is the second projection, unless the source is $0$-indexing, but this is the only thing that makes sense). The target morphism $t:G\times \Omega\to \Omega$ is the action map, $(g,m)\mapsto gm$. The identity morphism, composition, and inversion are left unspecified, but they are $i(m) = (1,m)$, where $1$ is the identity in $G$, $\mu : G\times G\times \Omega \to G\times \Omega$ is $\mu(g,h,m)=(gh,m)$, and $\tau(g,m) = (g^{-1},gm)$.
A more intuitive interpretation of the same groupoid
I think it's easier to say that the groupoid of $\Omega$, sometimes written as $B\Omega$, has objects $\Omega$ and if $x,y\in\Omega$, then the morphisms are given by $$B\Omega(x,y) = \{g\in G: gx=y \}.$$ Then it's pretty intuitive how identities and composition work and that this forms a groupoid.
Fibered categories?
From your question it seems that the context here has to do with fibered categories.
The fibered category in this context is given in the first sentence
My assumption is that this category has objects pairs $(G,M)$ with $G$ a group, $M$ a $G$-set, and morphisms $(\phi,f) : (G,M)\to (H,N)$ pairs with $\phi:G\to H$ a homomorphism of groups and $f:M\to N$ a function satisfying $f(gm) = \phi(g)f(m)$.
Then this category is fibered over the category of groups, with projection $(G,M)\mapsto G$.
There's not enough context for me to relate the first sentence to the second sentence. The second sentence appears to be going into some specifics about $G$-sets before returning to the category of all $G$-sets for all groups later.