= Solution
A $d$-dimensional <embedded submanifold> $F\subseteq M^m$ has the subspace topology and, near each of its points, a <manifold chart> identifying it with a coordinate slice $\mathbb R^d\times\{0\}$. These <slice charts for an embedded submanifold> give its smooth structure, and its inclusion is a <smooth embedding>.
A <regular value> $q\in N^n$ of a <smooth map> $f:M^m\to N^n$ is one for which $df_p:T_pM\to T_qN$ is <surjective> at every $p\in f^{-1}(q)$. If the preimage is nonempty this forces $m\geq n$.
The <inverse function theorem> used here says that if $G:U\subseteq\mathbb R^m\to\mathbb R^m$ is smooth and $DG_p$ is an <invertible matrix>, then its restriction to a suitable neighbourhood of $p$ is a <diffeomorphism> onto an open neighbourhood of $G(p)$, with smooth inverse.
To prove the <regular level set theorem>, fix $p\in f^{-1}(q)$ and choose coordinates centred at $p,q$. The coordinate representative $\widetilde f$ has a rank-$n$ <Jacobian matrix>. After permuting domain coordinates, its first $n$ columns form an <invertible matrix>. Define
$$
G(x)=\bigl(\widetilde f^1(x),\ldots,\widetilde f^n(x),x^{n+1},\ldots,x^m\bigr).
$$
Its <Jacobian matrix> is block triangular with invertible diagonal blocks, so the <inverse function theorem> makes $G$ a local coordinate change. In these coordinates, $f^{-1}(q)$ is precisely the slice with first $n$ coordinates zero. Thus it is an <embedded submanifold> of dimension $m-n$, with
$$
\boxed{T_p f^{-1}(q)=\ker df_p}.
$$
An empty preimage is harmless; the local assertion is then vacuous.
Three <equivalent formulations of orientability of a smooth manifold> are: an <oriented atlas> whose transition <Jacobian determinants> are positive; a smoothly varying <orientation of a vector space> on each <tangent space>; and a nowhere-vanishing smooth top-degree <differential form>. An oriented atlas gives compatible oriented coordinate frames. A <partition of unity> glues their positive local top forms, and a nonzero top form declares a frame positive precisely when it evaluates positively on that frame. These constructions prove the equivalence.
For the <orientability of a regular fibre>, write $F=f^{-1}(q)$. The derivative gives a <short exact sequence> of <vector bundles>
$$
0\longrightarrow TF\longrightarrow TM|_F
\xrightarrow{df}F\times T_qN\longrightarrow0.
$$
This is the reason the <normal bundle of a regular fibre is trivial>. Choose any fixed positive basis $b_1,\ldots,b_n$ of $T_qN$ and an ambient orientation on $M$. Declare a frame $v_1,\ldots,v_{m-n}$ of $T_pF$ positive when
$$
\widetilde b_1,\ldots,\widetilde b_n,v_1,\ldots,v_{m-n}
$$
is positive in $T_pM$, where $df_p(\widetilde b_i)=b_i$. Changing a lift adds a tangent vector and leaves the <determinant> sign unchanged. Local smooth lifts exist because $df$ is surjective, so this rule varies smoothly and agrees on overlaps. It gives a global orientation on $F$. Only an orientation of the single vector space $T_qN$ was chosen; $N$ itself need not be orientable.
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