Update: After depressingly many hours I realized that the task of bounding the integral is quite simple and in all likelihood, there is no such $u$-substitution conundrum as I have discussed here (you just perform integration by parts to the expression $\frac{-2\pi i f'(t)e^{-2\pi i f(t)}}{-2\pi i f'(t)}$). It is more likely that the combination of a different-ish way of writing mathematics + the conciseness of the author made me misinterpret the exact steps the author had in mind. I will, however, leave this question open because I have discussed here is still not clear to me: If you would perform the $u$-substitution, in what if in any way could you "move forward" with the integrand $\frac{1}{f'(u^{-1}(u))}$.
Let $f\in C^2([0,1])$ such that $|f'| > r$ for some $r > 0$ on $[0,1]$ and consider the integral
$$I = \int_0^1 e^{-2\pi i f(t)}dt$$
A proof I am reading wants to bound $|I|$ by performing such u-substitution that $du = e^{-2\pi i f(t)}f'(t)dt$ and integrating by parts to the function $\frac{1}{f'(t)}$. The proof does not give any details beyond the closing remarks on the tasked bound, so I am trying to replicate the entire process.
So what I have figured so was is that since $f'$ is outside some neighborhood of zero throughout $[0,1]$, we can safely write
$$I = \int_0^1\frac{-2\pi if'(t)}{-2\pi if'(t)}e^{-2\pi i f(t)}dt$$
The given $du$ suggest that
$$u(t) := \frac{e^{-2\pi i f(t)}}{-2\pi i}$$
so that
$$I = \int_{u(0)}^{u(1)}\frac{-2\pi i}{f'(t)}du$$
(there's likely a slight caveat here regarding integration over complex domain, which I am not sure about but will postpone to a later question).
Troubling part: At this point I am unfortunately a bit lost regarding two things:
What should the argument of $f'$ be after the $u$-substitution? 100% of the times I have had to use any substitution have been such that everything works out nicely in the end. Since after the substitution we are integrating over $u$'s image of $[0,1]$, wouldn't we have something like $f'(u^{-1}(u))$ instead of $f'(t)$ in $I$. That is,
$$I = \int_{u(0)}^{u(1)}\frac{-2\pi i}{f'(u^{-1}(u))}du$$
assuming that $u$ is invertible. And if so, how would you precisely do the integration by parts on the integral $I$? I get that you would integrate the constant function $1$ to be the monomial $u$, but my brain just shuts down when I try to work with $\frac{1}{f'(u^{-1}(u))}$ w.r.t. $u$.