Normal bundle of a regular fibre is trivial ($TM{|}_F/TF\cong F\times T_{q}N$)

For a regular value $q$ of a smooth map $f:M\to N$, let $F=f^{-1}(q)$. The differential of a smooth map identifies the quotient vector bundle $TM{|}_F/TF$ with the fixed target tangent space $T_{q}N$ over every point of $F$. After choosing a Riemannian metric, that quotient identifies with the normal bundle. Hence the normal bundle is trivial even when the target manifold has no global orientation.

Choose an orientation of a vector space on $T_{q}N$. The quotient identification for a regular value gives a smoothly oriented normal bundle along $F=f^{-1}(q)$. An oriented normal frame followed by a frame of $TF$ defines the latter's orientation through the ambient orientation. Changing normal lifts adds only tangent components, so the determinant sign is unchanged. No orientation on all of $N$ is required.