A closed submanifold $S\subset M$ of codimension $k$ is the inverse image of a regular value of a smooth map $f:M\rightarrow S^k$ if and only if it has trivial normal bundle.
One implication is explained in evgeniamerkulova's answer (and still holds with $S^k$ replaced by any other manifold of dimension $k$).
For the other one, pick a tubular neighbourhood $U$ of $S$ in $M$. Then $U$ is diffeomorphic to $S\times\mathbb{R}^k$, because $S$ has trivial normal bundle. You can now define a map $U\rightarrow \mathbb{R}^k$ by $(x,v)\mapsto v$ and extend it to $M$ by mapping the complement of $U$ to infinity. You can then approximate the resulting map by a smooth map that has $0$ as a regular value and whose zero set is $S$. (This is exercise 4.6.5 in Hirsch's Differential Topology).
Following the proof given in Milnor's Topology From A Differentiable Viewpoint:
$\require{AMScd}$
$\begin{CD}
x \in U \subseteq M @>F>> c \in V \subseteq N \\
@V\phi VV @VV\psi V \\
\phi\left(U\right) \subseteq H^{m} @>>\psi F \phi^{-1}> \psi\left(c\right) \in \psi\left(V\right)
\end{CD}$
Because $c \in N$ is a regular value of $F$, for every $x \in F^{-1}\left(c\right) \subseteq M$, there are charts $\left(U,\phi\right)$ at $x$ in $M$ and $\left(V,\psi\right)$ at $c$ in $N$ such that $\psi F \phi^{-1}: \phi\left(U\right) \subseteq H^{m} \to \psi\left(V\right) \subseteq \mathbb{R}^{n}$ is smooth, and has a regular value at $\psi\left(c\right)$.
$\begin{CD}
\phi\left(U\right) \subset W \subseteq \mathbb{R}^{m} @>G>> \psi\left(c\right) \in \mathbb{R}^{n}
\end{CD}$
Let $W$ be an open subset of $\mathbb{R}^{m}$ such that $W \cap H^{m} = \phi\left(U\right)$; and let $G:W\to\mathbb{R}^{n}$ be the smooth extension of $\psi F \phi^{-1}$ over $W$. Now, we can always choose $W$ small enough so that $G^{-1}\left(\psi\left(c\right)\right)$ does not contain any critical points (Sard's lemma; $\mathbb{R}^{m}$ is regular). Thus, $\psi\left(c\right)$ is a regular value of $G$; and by preimage theorem (for smooth manifolds), $Z := G^{-1}\left(\psi\left(c\right)\right)$ is a (m-n)-dimensional submanifold of $\mathbb{R}^{m}$.
Furthermore, since $G$ is constant over $Z$, $T_{a}Z \subseteq ker\left\{DG\left(a\right)\right\}$ for every $a \in Z$. But $DG\left(a\right):\mathbb{R}^{m}\to\mathbb{R}^{n}$ is surjective, and the rank-nullity theorem implies $dim \left(ker\left\{DG\left(a\right)\right\}\right) =$ (m-n). Thus, $T_{a}Z = ker\left\{DG\left(a\right)\right\}$.
Now, define $\pi: Z \subseteq \mathbb{R}^{m} \to \mathbb{R}$ as $\left(x_{1},\ldots,x_{m}\right) \mapsto x_{m}$.
To show that $0 \in \mathbb{R}$ is a regular value of $\pi$:
Suppose otherwise. That is, suppose $\exists$ $a \in Z \cap \partial H^{m}$ such that $d\pi_{a}:T_{a}Z \to \mathbb{R}$ is not surjective. Then, $ker \left\{d\pi_{a}\right\}$ $=$ $T_{a}Z$ $=$ $ker \left\{DG(a)\right\}$. But, we know that $ker \left\{d\pi_{a}\right\} \subseteq \mathbb{R}^{m-1} \times \left\{0\right\} = \partial H^{m}$. Thus, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ if $0$ is not a regular value for $\pi$.
Now, since $c \in N$ is a regular value of $F|_{\partial M}$ as well, arguing as before, we can show that $\bar{G} := G|_{W\cap\partial H^{m}}$ has a regular value at $\psi\left(c\right)$. That is, for every $a \in Z\cap\partial H^{m}$, $D\bar{G}\left(a\right): \mathbb{R}^{m-1} \to \mathbb{R}^{n}$ is surjective; and by rank-nullity theorem, $dim \left(ker\left\{D\bar{G}\left(a\right)\right\}\right) = $ (m-n-1)
Finally, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ implies $ker \left\{DG(a)\right\} = ker \left\{D\bar{G}(a)\right\}$, which is clearly false (dimension mismatch). Hence, $0$ must be a regular value for $\pi$.
Since $0 \in \mathbb{R}$ is a regular value for $\pi$, $\left\{z \in Z | \pi\left(z\right) \geq 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right)\right)$ is a manifold with boundary $\left\{z \in Z | \pi\left(z\right) = 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right) \cap \partial M \right)$.
$\phi$ being a diffeomorphism, $U \cap F^{-1}\left(c\right)$ is a manifold with boundary $U \cap F^{-1}\left(c\right) \cap \partial M$. Observing that this is true for every $x \in F^{-}\left(c\right)$ completes the proof.
Best Answer
$f^{-1}(c)$ is orientable if and only if $f^{-1}(c-\epsilon,c+\epsilon)$ is orientable, assuming sufficiently small $\epsilon$.
If $M$ itself is orientable then that is a sufficient condition, because $f^{-1}(c-\epsilon,c+\epsilon)$ is an open subset and therefore orientable.
I don't think there's much more you can say than that.