$(\Rightarrow)$ 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
If $\det\bigl(\operatorname{Jac}(\tau)\bigr) < 0$, you can re-orient one of your charts (for example, by precomposing with a linear diffeomorphism that multplies the first coordinate by $-1$).