$(\Rightarrow)$ 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 answer is given by Kajelad in the comment
You have proved that $M$ is orientable $\Rightarrow TM$ is orientable. Now we prove the opposite direction.
Assume that $TM$ is orientable. Then there is a family of open cover $\{U_i\}_{i\in \Lambda}$ of $M$ and for each $i\in \Lambda$, a local trivialization
$$\varphi_i : TU_i \to U_i \times \mathbb R^n$$
so that for all $i, j$ with $U_i \cap U_j \neq \emptyset$, the transition function
$$ g_{ij} : U_i \cap U_j \to \operatorname{GL}_n(\mathbb R)$$
has $\det g_{ij} >0$.
By shrinking to smaller open sets if necessary, we assume that each $U_i$ is a coordinates neighborhood. That is, there is $\psi_i : U_i \to \psi (U_i) \subset \mathbb R^n$ which is a local chart. By composing with a reflection of $\mathbb R^n$ if necessary, we assume that in $U_i$, both
$$\left\{ \frac{\partial }{\partial x_1}, \cdots, \frac{\partial }{\partial x_n}\right\}, \{ \varphi^{-1}_i e_1, \cdots, \varphi^{-1}_i e_n\}$$
have the same orientation. Here $\{e_1, \cdots, e_n\}$ are the standard basis of $\mathbb R^n$. This is the same as saying that for all $x\in U$, the linear map $L_i (x)$ defined by the composition
$$ \mathbb R^n \cong T_{\psi(x)} \psi_i (U_i))\overset{(\psi^{-1}_i)_*}{\to} T_xU \overset{\varphi_i|_{T_xU_i}}{\to} \mathbb R^n$$
has positive determinant.
Now we check that $M$ is orientable: whenever $U_i \cap U_j$ is non-empty, let $x\in U_i$. Then one need to check $J_{ij}:= J(\psi_i \circ \psi_j):\mathbb R^n \to \mathbb R^n$ has positive determinant, but this is true since
\begin{align}
J(\psi_i \circ \psi_j^{-1}) &= (\psi_i)_* \circ (\psi_j^{-1})_* \\
&= (L_i^{-1} \circ \varphi_i) \circ (\varphi_j^{-1} \circ L_j)\\
&= L_i^{-1} \circ g_{ji} \circ L_j.
\end{align}
Best Answer
Your assumption of a trivialization $f: M\times \Bbb R^n \to TM$ yields, for each $p\in M$, a linear isomorphism $f_p:\Bbb R^n \to T_pM$ given by the composition of $f|_{\{p\}\times \Bbb R^n}$ with the identification $\{p\}\times \Bbb R^n \approx \Bbb R^n$.
Now let $\cal A$ be the maximal atlas for $M$. For any chart $(\phi_j,U_j) \in \cal A$, say that $\phi_j$ is "good" if for all $p\in U_j$, $$\det\left[(D_p \phi_j) \circ f_p\right] > 0,$$ and that $\phi_j$ is "bad" otherwise. Now remove all of the "bad" charts from $\cal A$ to get a collection $\cal{B}$. Since any good chart can be obtained from a bad chart (and vice versa) by multiplying one of the coordinate functions by $-1$, $\cal{B}$ is an (nonmaximal) atlas.
You can check that the atlas $\cal{B}$ satisfies your specified orientation-preserving criterion (use the chain rule).
Edit (more details): for any pair of "good" charts, by the chain rule it follows that $$D_{\phi_j(p)}(\phi_i\circ \phi_j)^{-1} = \left[(D_{p}\phi_i)\circ f_{p}\right] \circ \left[(D_{p}\phi_j)\circ f_{p}\right]^{-1}, $$ which has positive determinant as desired.