$(\Rightarrow)$ 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:
$$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$
Best Answer
If $M\times N$ is orientable, any open submanifold is orientable. We can pick an open subset $U\subset N$ diffeomorphic to $\mathbb R^n$, and $M\times U\equiv M\times\mathbb R^n$ is orientable. By induction it is enough to see that if $M\times\mathbb R$ is orientable, then $M$ is orientable. Pick any open cover $\{W_i\}$ of $M$ such that there are diffeomorphisms $\varphi_i:\mathbb R^m\to W_i$. The cover ${\mathcal A}=\{W_i\times\mathbb R\}$ is an atlas with parametrizations $\psi_i=\varphi_i\times Id:\mathbb R^{n+1}\to W_i\times\mathbb R$. Then, if needed we can modify each $\psi_i$ by changing the sign of the first variable in $\mathbb R^{n+1}$ to make it compatible with a fixed orientation in $M\times\mathbb R$. This changes correspondingly the $\varphi_i$. Thus ${\mathcal A}$ is positive and we have $$ J(\psi_j^{-1}\circ\psi_i)=\begin{pmatrix} J(\varphi_j^{-1}\circ\varphi_i)&0\\0&I \end{pmatrix}, $$ hence $J(\varphi_j^{-1}\circ\varphi_i)=\det J(\psi_j^{-1}\circ\psi_i)>0$. Thus the $\varphi_i$'s are a positive atlas of $M$. We are done.