Equivalent Characterizations of Smoothness. 
Definition. Let $M$ and $N$ be smooth manifolds, and let $F\colon M\to N$ be any map. We say that $F$ is a smooth map if for every $p\in M$, there exists smooth chart $\big(U,\varphi\big)$ containing $p$ and $\big(V,\psi\big)$ containing $F(p)$ such that $F(U)\subseteq V$ and the composite map $\psi\circ F\circ \varphi^{-1}\colon \varphi(U)\to\psi(V)$ is smooth.

I have to show the following proposition, but many points are not clear to me.

Proposition. Suppose $M$ and $N$ are smooth manifolds with or without boundary, and $F\colon M\to N$ is a map. Then $F$ is smooth if and only if either of the following condition is satisfied:
$(a)$ For every $p\in M$, there exist smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $U\cap F^{-1}(V)$ is open in $M$ and the composite map $\psi\circ F\circ\varphi^{-1}$ is smooth from $\varphi(U\cap F^{-1}(V))$ to $\psi\big(V\big)$.
$(b)$ $F$ is continuous and there exists amooth atlases $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ for $M$ and $N$ respectevely such that for each $\alpha$ and $\beta$, $\psi_\beta\circ F\circ \varphi_\alpha^{-1}$ is a smooth map from $\varphi_\alpha(U_\alpha\cap F^{-1}(V_\beta))$ to $\psi(V_\beta)$.

Proof. I would like to proceed, hoping that it is the simplest way as follows: first prove that $F$ is smooth iff $(a)$ holds, and after showing that $(a)$ holds iff $(b)$ holds.
F is smooth iff $(a)$ holds
($\Rightarrow$) If $F$ is smooth we have that $F(U)\subseteq V$, then $U\subseteq F^{-1}(V)$. Therefore $U\cap F^{-1}(V)=U$ that is open.
$(\Rightarrow)$ If $(a)$ holds for every $p\in M$, there exist smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $U\cap F^{-1}(V)$ is open in $M$ and the composite map $\psi\circ F\circ\varphi^{-1}$ is smooth from $\varphi(U\cap F^{-1}(V))$ to $\psi\big(V\big)$. It only remains to show that $F(U)\subseteq V$, now if $\tilde{p}\in U$ then there exists a smooth chart $(\tilde{V},\tilde{\psi})$ containing $F(\tilde{p})$

Question 1. How can I prove that $F\big(\tilde{q}\big)\in V$?

(a)$\iff$(b)
$(b)\Rightarrow (a)$ it seems obvious.

Question 2. I have no idea how to show that $(a)\Rightarrow (b)$. Could you give me a hints?

Thanks!
 A: 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
$$\underbrace{\psi_\alpha\circ\psi_p^{-1}}_{\substack{\text{smooth because of the} \\ \text{coherence of the charts}}}
\circ\underbrace{\psi_p\circ F\circ \varphi_p^{-1}}_{\substack{\text{smooth by assumption} }}.$$
