The reason we have charts on a topological manifold is so that if we pick a chart $U \to \mathbb{R}^n$ that is a continuous function, we can say that the topological space $U$ is homeomorphic to the topological space $\mathbb{R}^n$, so any questions about the topology of $U$ can be moved to questions about the topology of Euclidean space, which we understand very well.
For differentiable manifolds, we want charts that make $U$ diffeomorphic to $\mathbb{R}^n$, so that any questions about the differentiable structure on $U$ can be moved to questions about the differentiable structure of Euclidean space, which we understand well.
For smooth manifolds, we want $U$ to be isomorphic to $\mathbb{R}^n$ in the appropriate sense as well.
This is why we don't consider all charts on a differentiable manifold or a smooth manifold.
Of course, for this to make sense, we need a way to be able to define what it means for a map to be diffeomorphic. We know how to define diffeomorphisms of subspaces of Euclidean space. The transition map idea is a device to let us take that definition and use it to define what a diffeomorphism is from a subspace of our manifold. (and happily, that turns out to be all we need to make the definition work)
It may be a fun exercise to show the analogous fact for topological manifolds. Suppose you didn't even bother defining a topology on the set $M$: you just had an atlas of charts that are just ordinary bijective functions. Try using the atlas to define a topology on $M$. You'll find that you need the transition maps to be continuous, and that's all you need.
(EDIT: you also need the domain and image of the transition maps to be open)
Question 1: $F(U)\subset V$ is not always true. But you can take a new chart of the form $(U^\prime,\varphi)$ which has this property (take $U^\prime=U\cap F^{-1}(V)$).
Question 2: If you want to do it without "$a\Leftrightarrow F \text{ smooth}$", which I guess you do:
For $p\in M$ take $(U_p,\varphi_p)$ and $(V_p,\psi_p)$ some charts around $p$ and $F(p)$ respectively. If $U_p^\prime=U_p\cap F^{-1}(V_p)$ then $\{ U_p^\prime\}_p$ is an open cover for $M$ such that $F_{\vert U_p^\prime}$ is continuous, so $F$ is continuous.
Take the atlas $\{(U_p^\prime,\varphi_p)\}_p$ for $M$. Unfortunately $\{(V_p,\psi_p)\}$ may not be an atlas for $N$ because the domain of the charts may not cover $N$ is $F$ is not surjective. You can take the original atlas of $N$ instead, which we write $\{(V_\alpha,\psi_\alpha)\}$. Then for $p$ and $\alpha$ the map
$$\psi_\alpha\circ F\circ \varphi_p^{-1}:\varphi_p(U_p^\prime\cap F^{-1}( V_\alpha))\longrightarrow\psi(V_\alpha)$$
has domain $\varphi_p(U_p^\prime\cap F^{-1}( V_\alpha))=\varphi_p(U_p\cap F^{-1}(V_p\cap V_\alpha))$ so it is just the composition
$$\underbrace{\psi_\alpha\circ\psi_p^{-1}}_{\substack{\text{smooth because of the} \\ \text{coherence of the charts}}}
\circ\underbrace{\psi_p\circ F\circ \varphi_p^{-1}}_{\substack{\text{smooth by assumption} }}.$$
Best Answer
By definition, a manifold is a set endowed with a maximal atlas, so that you will never have to consider multiple atlases for a manifold and none of the problems you pointed out exist.