$(\Rightarrow)$ 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:
$$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$
Here's a fairly detailed sketch:
Let $M$ be an $n$-manifold. Suppose, contrapositively, that $M$ is orientable, and fix a maximal oriented atlas, i.e., a maximal atlas for which all the transition maps have Jacobian with positive determinant.
Lemma 1: If $(U, \phi)$ is a chart and $U$ is connected, then $(U, \phi)$ is either compatible with the orientation, or anti-compatible with the orientation.
Proof: Let $(V, \psi)$ be an arbitrary oriented chart with $U \cap V$ non-empty. The sign of the Jacobian of $\psi \circ \phi^{-1}$ in $\phi(U \cap V)$ is independent of $\psi$ because $(V, \psi)$ is selected from an oriented atlas.
Let $U^{+}$ denote the set of points of $U$ for which the Jacobian $\psi \circ \phi^{-1}$ is positive, and let $U^{-}$ denote the set of points of $U$ for which the Jacobian $\psi \circ \phi^{-1}$ is negative. The sets $U^{\pm}$ are obviously disjoint, each is open, and their union is $U$. Since $U$ is connected, one set is empty: Either $U = U^{+}$ or $U = U^{-}$.
Lemma 2: If $(U, \phi)$ is an anti-compatible chart with component functions $(\phi^{1}, \dots, \phi^{n})$, then the chart $(U, \bar{\phi})$ defined by $\bar{\phi} = (\phi^{1}, \dots, \phi^{n-1}, -\phi^{n})$ is compatible, and conversely.
Proof: The linear transformation $T(x^{1}, \dots, x^{n}) = (x^{1}, \dots, x^{n-1}, -x^{n})$ has determinant $-1$.
Let $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ be arbitrary charts with $U_{a}$ and $U_{b}$ connected, and with $U_{a} \cap U_{b}$ non-empty. If necessary, replace $\phi_{a}$ by $\bar{\phi}_{a}$ to get a compatible chart, and similarly for $\phi_{b}$. Since each chart is compatible with the oriented atlas, the transition map has positive Jacobian. It follows that the transition map $\phi_{ab}$ between the original charts has Jacobian of constant sign, i.e., either preserves orientation or reverses orientation.
Contrapositively, if there exist connected, overlapping charts $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ whose transition map $\phi_{ab}$ neither preserves nor reverses orientation, then $M$ is not orientable.
Best Answer
One definition (#1) of the orientation of a manifold $X$ is a choice of orientation on $T_xX$ for all $x \in X$ which varies continuously with $x$. So an orientation on a manifold by definition corresponds to choosing orientations on $T_xX$ continuously for all $x \in X$.
Aside on orientations of vector spaces
(#1') Let $V$ be a real $n$-dimensional vector space, and $(v_1, \dots, v_n)$ and $(v_1', \dots, v_n')$ be bases of $V$. Then $v_i'= \sum_{j=1}^n A_{ij}v_j$ for $A \in \mathrm{GL}(n, \mathbb{R})$. Now define an equivalence relation on bases of $V$ by $(v_1, \dots, v_n) \sim (v_1', \dots, v_n')$ if and only if $\det A > 0$. Then an orientation of $V$ is a choice of equivalence class $[ v_1, \dots, v_n ]$.
(#2') There is a second notion of orientations of vector spaces. Let $V$ be as above. Then $\bigwedge^n V \cong \mathbb{R}$ and we have $0 \neq v_1 \wedge \cdots \wedge v_n \in \bigwedge^n V$. By what we've said, $\bigwedge^n V \setminus 0 \cong (-\infty, 0) \cup (0, \infty)$ has two connected components.
It is a fact that if $(v_1, \dots, v_n)$ and $(v_1', \dots, v_n')$ are bases of $V$ with $v_i'= \sum_{j=1}^n A_{ij}v_j$ then $$ v_1' \wedge \cdots \wedge v_n' = \det A \cdot v_1 \wedge \cdots \wedge v_n . $$
So define an equivalence relation $(v_1, \dots, v_n) \sim (v_1', \dots, v_n')$ if and only if $ v_1' \wedge \cdots \wedge v_n'$ and $v_1 \wedge \cdots \wedge v_n $ lie in the same connected component of $\bigwedge^n V \setminus 0$.
You may ask how the definition of orientation of a manifold that I've given above and the definition you've given in terms of charts are equivalent.
Firstly, as we've said in (#2') above, one can interpret an orientation on a vector space $V$ as choosing a connected component of $\bigwedge^n V \setminus 0$. Applying this definition of orientations of vector spaces to the vector spaces $T_xX$ gives a second definition for the orientation of a manifold:
#2: An orientation of an $n$-manifold $X$ is an equivalence class $[ \omega ]$ of non-vanishing top forms $\omega \in \Omega^n(X)$ where $\omega$ is equivalent to $\omega'$ if and only if $\omega'=f \circ \omega$ for a smooth function $f : X \to (0 , \infty)$.
We finally come to the third definition (your definition) of the orientation of a manifold:
#3: We first define oriented charts. Let $X$ be an $n$-manifold with orientation $[ \omega ]$. Let $(U, \phi)$ be a chart on $X$. We call $(U, \phi)$ oriented if $\phi^*(\omega) = f \cdot dx_1 \wedge \cdots \wedge dx_n$ where $f : U \to \mathbb{R}$ and $f >0$. we can find an atlas $\mathcal{A} = \{ (U_i, \phi_i) \mid i \in I \}$ for $X$ consisting of only oriented charts. Call $\mathcal{A}$ an oriented atlas. For two such charts on $X$ with local coordinates $(x_1, \dots, x_n)$, $(y_1, \dots, y_n)$ we have $\det \left( \frac{ \partial y_i}{\partial x_j} \right)_{i, j=1}^n > 0$ on the overlap. Then we can define an oriented manifold to be a manifold with oriented atlas.
Hopefully this chain of definitions sufficiently links together the way of thinking about orientations of manifolds via charts and the way of thinking about orientations of manifolds as orientations of their tangent spaces.