Notice that $g(t), 0\le t\le 2\pi$, is a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $|f'(t)|$. By assumption, $a\le |f'|\le b$.

For convenience, let's discuss the case $x=0$.
I claim that the inequality may be viewed as the condition that we will still be able to return $g(0)$, given the restrictions on the curvature.

The comparison curve $e^{it}$ is the unit circle, and if we consider the points corresponding to $t=0$, $t=x$ and the center of this circle, we obtain an isosceles triangle with base of length $|e^{ix}-1|$ and legs of length $1$.

Now compare this with the corresponding points $g(x)$, $g(0)=0$. The maximum curvature we are allowed to use is $b$, so we have to be able to return from $g(x)$ to $g(0)$ along an arc of length $2\pi -x$. To minimize the length needed, we should make the curvature large, at least initially, but we are not allowed to go beyond $b$.

This suggests a corresponding isosceles triangle with base $|g(x)-g(0)|$ and legs of length $1/b$. If your (second) inequality holds with equality, this triangle is similar to the one from above, and we are just about able to return to $g(0)$. If $|g(x)-g(0)|$ became larger still, we would run into a contradiction.

The first inequality can be proved similarly, of course.