$(\Rightarrow)$ If $M^m\times N^n$ is orientable, there is a volume form $\eta\in\Omega^{n+m}(M\times N)$. For any fixed point $q\in N$, take a basis $\{w_1,...,w_n\}$ for $T_qN$. Now define $\omega\in\Omega^m(M)$ by:
$$\omega(X_1,...,X_m):=\eta_{(\cdot,q)}\left(X_1,...,X_m,w_1,...,w_n\right)$$

We will prove that $\omega$ is a volume form. For a fixed $p$, take a basis $\{v_1,...,v_m\}$ for $T_pM$. Using the identification $T_{(p,q)}M\times N\equiv T_pM\oplus T_qN$, then $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_{(p,q)}M\times N$. Since $\eta$ is a volume form, $\omega_p(v_1,...,v_m)=\eta_{(p,q)}(v_1,...,v_m,w_1,...,w_n)\neq 0$, which means $\omega$ is a volume form, so $M$ is orientable. By a similar argument, $N$ is orientable.$_\blacksquare$

$(\Leftarrow)$ If $M,N$ are orientable, there are volume forms $\omega\in\Omega^m(M), \sigma\in\Omega^n(N)$. For the natural projections $\pi_M:M\times N\to M$ and $\pi_N:M\times N\to N$, define:
$$\eta:=\pi_M^*(\omega)\wedge \pi_N^*(\sigma)\in\Omega^{n+m}(M\times N)$$

We will prove $\eta$ is a volume form. For a fixed $x=(p,q)\in M\times N$, take basis $\{v_1,...,v_n\}$ for $T_pM$ and $\{w_1,...,w_n\}$ for $T_qN$. Because $\omega$ and $\sigma$ are volume forms, we have $\omega_p(v_1,...,v_m)\neq 0$ and $\sigma_q(w_1,...,w_n)\neq 0$. Since $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_x(M\times N)$, it's enough to check that $\eta_x(v_1,...,v_m,w_1,...,w_n)\neq 0$. Indeed, noticing that $(d\pi_M)_x(v_i)=v_i$, $(d\pi_M)_x(w_j)=0$, $(d\pi_N)_x(v_i)=0$ and $(d\pi_N)_x(w_j)=w_j$, we have:
$$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$

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)

## Best Answer

You can use the chart $$ \mathbf x(u,v) = \left(\frac{u}{\sqrt{u^2+v^2}}, \frac{v}{\sqrt{u^2+v^2}}, \log \sqrt{u^2+v^2} \right)\,,$$ defined on $\mathbf R^2 \setminus \{(0,0)\}$. The formula is simpler in polar coordinates, $$ \mathbf x(r,\theta) = \left( \cos \theta, \sin \theta, \log r\right)\,. $$