It looks like, in all dimensions, $\sigma_+$ preserves orientation and $\sigma_-$ reverses it. Here is the case of $S^2$ explicitly worked out:
For a vector space $V$ of dimension $n$, an orientation will be an element of
$$ \{(b_1,\dots,b_n)\ |\ V=\langle b_1,\dots, b_n\rangle \}/GL^+(V) \cong \mathbb{Z}/2$$
and given oriented vector spaces $(V,o)$ and $(V',o')$, a linear isomorphism $L\colon V\to W$ will be orientation preserving if $L(o)=o'$, and orientation reversing if $L(o)=-o'$.
For the sake of clarity, we will write $\mathbb{R}^3=\langle e_1,e_2,e_3\rangle$ and $\mathbb{R}^2=\langle e_1',e_2'\rangle$, and we will give these vector spaces the natural orientations.
Explicitly, the stereographic projections are given by $$\sigma_+(x,y,z)=(\frac{x}{1+z},\frac{y}{1+z})\text{ and }\sigma_-(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z})$$ for $(x,y,z)$ in the appropriate domains.
Consider specifically the point $e_1=(1,0,0)$, which is in the domains of both functions. An orientation of $T_{e_1}S^2\cong \langle e_2,e_3\rangle$ is an ordered basis with the property that concatenating with the outward-pointing normal vector $e_1$ gives a positively oriented basis for $\mathbb{R}^3$. (In this case it doesn't matter if the concatenation happens at the beginning or end, because they differ by an even number of transpositions.) The preferred orientation for $\mathbb{R}^3$ is $[e_1,e_2,e_3]$, so the orientation on $T_{e_1}S^2\cong \langle e_2,e_3\rangle$ is given by $[e_2,e_3]$.
Stereographic projection sends the standard great circle $S^{1}\subset S^2$ to the unit circle $S^{1}\subset\mathbb{R}^2=\langle e_1',e_2'\rangle$, and in particular $e_1$ is sent to $e_1'$. The tangent space $T_{e_1'}\mathbb{R}^2$ is naturally isomorphic to $\langle e_1',e_2'\rangle$, and the preferred orientation is $[e_1',e_2']$.
In order to compute the derivative of $\sigma_{\pm}$ at $e_1$, consider the two paths
$$ \gamma_2(t)=(cos(t),sin(t),0)\text{ and }\gamma_3(t)=(cos(t),0,sin(t))$$
Then $\frac{d}{dt}\gamma_i(t)|_{t=0}=e_i$, i.e. the derivatives form a basis for the tangent space at $e_1$. Compute
$$\frac{d}{dt}\sigma_-\big(\gamma_2(t)\big)|_{t=0} = \frac{d}{dt}\big(cos(t),sin(t)\big)|_{t=0}=(0,1)=e_2'$$
and
$$\frac{d}{dt}\sigma_-\big(\gamma_3(t)\big)|_{t=0} = \frac{d}{dt}\left(\frac{cos(t)}{1-sin(t)},0\right)|_{t=0}=(1,0)=e_1'$$
Hence the derivative of $\sigma_-$ at $e_1$ takes the positively oriented basis $[e_2,e_3]$ to $[e_2',e_1']=-[e_1', e_2']$, and so $\sigma_-$ reverses orientation at $e_1$. That it reverses orientation at all points in its domain should follow from continuity of $\sigma_-$ and connectedness of its domain.
Similarly, one computes that the derivative of $\sigma_+$ at $e_1$ takes the basis $[e_2,e_3]$ to $[e_2',-e_1']=[e_1’,e_2’]$, and so it preserves orientation.
A similar computation also works in higher dimensions, but I wrote out the case of $S^2$ because it still illustrates the key idea and computation. For $S^{2n+1}$ one needs to be careful about defining an oriented basis for a tangent space, because moving a normal vector from the end of an ordered basis to the beginning is an odd permutation; however, if the convention is that the normal vector is added at the end, then again we should have $\sigma_-$ orientation reversing and $\sigma_+$ orientation preserving.
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
No, you can't. Intuitively, the mapping from non-empty regular levels to $N$ is a bijection, so a canonical orientation on the normal bundle to an arbitrary regular level induces an orientation on $N$.
The simplest example is probably to take $M$ to be a solid torus, $N$ a Möbius strip embedded in $M$ as a rotating segment in the disks perpendicular to the central circle, and $f:M \to N$ the orthogonal projection, which is clearly a submersion. Each level of $f$ is a line segment $S$ in $M$, whose normal bundle is (isomorphic to) $S \times \mathbf{R}^{2}$. But if we fix an orientation on the normal bundle of $S$ and try to extend by continuity, we run into trouble when we walk around the central circle.