 Normal bundle of a regular fibre is trivial

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.