Take $f(x)=k\log(x+1)$ where $k=2\pi/\log(2\pi+1)$, hence $f$ is a diffeomorphism. The maximum of the derivate is atained when $x=0$ and it is $k\approx 3.1644$. On the other hand, taking $x=\pi/3$, $y=\pi/3-1$ we obtain
$$ \frac{|g(x)-g(y)|}{|e^{\mathrm{i}x}-e^{\mathrm{i}y}|}> \frac{2}{0.2}=10.$$
Edit: As Chistian noted in a comment, with the function $f$ defined above the function $g$ don't satisfy $g(0)=g(2\pi)$.
Now, we can choose $f(x)=-x+2\pi$. For $g(2\pi)$ we get
$$g(2\pi)=\int_0^{2\pi}e^{-\mathrm{i}x+2\pi\mathrm{i}}=\int_0^{2\pi}e^{-\mathrm{i}x}=0.$$
The derivate is $-1$, and
$$ \frac{|g(x)-g(y)|}{|e^{\mathrm{i}x}-e^{\mathrm{i}y}|}=1.$$
This is not a surprise if we think the above quotient as
$$ \left|\frac{g(x)-g(y)}{x-y}\frac{x-y}{e^{\mathrm{i}x}-e^{\mathrm{i}y}}\right|.$$ If the estimate is true, then $\underline{f}\leq|g^\prime(x)|\leq \overline{f}$ for every $x\in(0,2\pi)$. But $|g^\prime(x)|=|f^\prime(x)|$. I will think about the case where $\underline{f}=\min |f^\prime|$ and $\overline{f}=\max |f^\prime|$.