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.
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.
Best Answer
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}