Preimage of tubular neighborhood Let $f:M' \to M$ be a map between smooth manifolds. Let $S \subset M$ be a submanifold, and let $T$ be a tubular neighborhood of $S$, ie. $T$ is diffeomorphic to the normal bundle of $S$ in $M$. If $f$ is transversal to $S$ and $T$, $f^{-1}(T)$ is a tubular neighborhood of $f^{-1}(S)$ provided $T$ is small enough.
This fact was used in P.69 of Bott, Tu, but I cannot come up with a proof of this. Ideally I would like to construct a diffeomorphism between $f^{-1}(T)$ and the normal bundle of $f^{-1}(S)$ directly, but I have not succeeded. I noticed the authors explicitly stated $T$ needs to be sufficiently small, this lead me to think maybe we have to approach this locally, perhaps allowing us to consider everything to be in Euclidean space, but I have no clear idea how this is done. Any help is appreciated!

Perhaps I'm missing your point, but the key issue is that when $f$ is transverse to $S$, not only is $f^{-1}(S)$ a submanifold of $M'$, but also one has 
$$N_{f^{-1}(S)/M'} = f^*N_{S/M}.$$ (Then it's just a matter of applying the tubular neighborhood theorem.) Indeed, the bundle map $df$ maps $N_{f^{-1}(S)/M'}$ isomorphically to $N_{S/M}$. (Although this is often not stated explicitly, $f\pitchfork S \iff df_x$ maps $T_xM'$ surjectively to $T_{f(x)}M\big/T_{f(x)}S \cong \big(N_{S/M}\big)_{f(x)}$ for each $x\in f^{-1}(S)$.)
