The group of isometries generated by $F$ is simply the set of powers of $F$ (i.e. $\{1, F, F^2, F^3, \dots\}$) and their inverses. Since $F$ is just rescaling by $2$, the group is just all rescalings by powers of $2$.
The quotient manifold $M \, / \, \Gamma$ is the set of equivalence classes of points in $M$ under the equivalence relation of being in the same orbit of $\Gamma$. By the description above, this means that $(x_1,y_1) \sim (x_2,y_2)$ if there is some integer $n$ so that $(x_2,y_2) = 2^n (x_1,y_1)$. So in the quotient, the points $(x,y)$, $(2x,2y)$, $(4x,4y)$, $(8x,8y)$, etc... all represent the same point. It can be identified by the annulus because any point can be multiplied by some $2^n$ to bring it to a point with a radius between 1 and 2. In other words, any equivalence class has a representative in this annulus.
The definition of properly discontinuous action can be found in a topology textbook. For example, see the wikipedia page: here.
If you're familiar with covering spaces, then there's an approach which should be useful for each of these:
Given a Riemannian manifold $(M,g)$, the universal cover $\widetilde{M}$ can be equipped with a metric $\widetilde{g}$ such that the covering map $\widetilde{M}\mapsto M$ coincides with the quotient map $\widetilde{M}\mapsto\widetilde{M}/\Gamma$ where $\Gamma$ is a discrete subgroup of isometries of $\widetilde{M}$. Using among other things the lifting property of universal coverings, you can show that the isometry groups are related by the following expression:
$$
\operatorname{Isom}(M)\cong N_{\operatorname{Isom}(\widetilde{M})}(\Gamma)/\Gamma
$$
(here $N$ denotes the normalizer)
Each of the listed manifolds are universally covered by $\mathbb{R}^2$, so they can all be described as quotients of the Euclidean plane by discrete subgroups of $E(2)$. Actually computing these will depend on which flat metrics you're interested in.
Best Answer
I'm going to consider the flat torus $\mathbb T = \mathbb R^2 / \Gamma$ for some plane lattice $$\Gamma_{e_1, e_2} = \{ z_1 e_1 + z_2 e_2 : z_1, z_2 \in \mathbb Z \}.$$ (Here $e_1, e_2$ is any basis of $\mathbb R^2$.) Since any isometry of the torus lifts to an isometry of $\mathbb R^2,$ we just need to determine which isometries of the plane descend to diffeomorphisms of $\mathbb T.$
A map $\phi: \mathbb R^2 \to \mathbb R^2$ descends iff it satisfies $$\phi(x+z) - \phi(x) \in \mathbb \Gamma \textrm{ for all } x \in \mathbb R^2, z \in \mathbb \Gamma.\tag 1 $$ Since the isometries of the plane can be written as $\phi(x) = Ax + c$ for $A \in O(2), c \in \mathbb R^2,$ this condition reduces to the fact that $Az = z$ for all $z,$ i.e. $A(\Gamma) \subset \Gamma.$ Finally, to have an inverse, $\phi^{-1}$ must also descend, and thus the condition is $A(\Gamma) = \Gamma;$ so $\phi(x) = Ax + c$ is an isometry of $\mathbb T$ iff the linear isometry $A$ is also an automorphism of the lattice $\Gamma.$
Note that two isometries induce identical maps on $\mathbb T$ if their constant terms differ by an element of $\Gamma,$ and composition of translations corresponds to addition of constant terms; so we can conclude that $$\mathrm{Isom}(\mathbb T) = (O(2)\cap \operatorname{Aut}(\Gamma)) \ltimes (\mathbb T, +). $$ (Compare to $\operatorname{Isom}(\mathbb R^2) =O(2) \ltimes (\mathbb R^2, +).$)
The nature of the normal subgroup $G = O(2)\cap \operatorname{Aut}(\Gamma)$ depends on the isometry class of the lattice (or equivalently on the shape of the fundamental domain):