Indeed you can use a substitution like $x = 4 \cos u$ to take the antiderivative of $\sqrt{16 - x^2}.$ I say an substitution "like" that, because $x = 4 \cos u$ is not a complete account of the substitution.
In a substitution into an antiderivative, you start with the original antiderivative, whose integrand is a function of some variable, say $x$; then for any particular value of $x$ you replace the integrand with a new integrand evaluated at a particular value of some other variable, say $u$. In this sense, $u$ is actually a function of $x,$ not the other way around, even though it is convenient (for the purpose of replacing occurrences of $x$ in the formula) to write the substitution with $x$ on one side of an equation.
If you are using a non-invertible function for the substitution you may have multiple choices of the value of $u$ at which you can evaluate the new integrand for any particular value of $x$; but it has to be the same value of $u$ in all parts of the integrand, and it has to remain the same value of $u$ during the remaining evaluation, in order for the substitution to make sense.
If, somewhere in the midst of evaluating the new integral,
you forget which value of $u$ you had in mind when you made the substitution, you are likely to get an incorrect result.
You have been shown how to identify a particular value of $u$ for each $x$ by restricting the choice of $u$ to $u \in [0, \pi].$
In effect, this says that $u = \arccos\left(\frac14 x\right).$
This also implies that $\lvert \sin u\rvert = \sin u$ at every possible value of $u.$
It also implies that $\sin u \geq 0.$
But it is actually possible for $\sin u$ to have a negative value.
You just have to set up your substitution to make this possible.
An example of such a substitution is $u = -\arccos\left(\frac14 x\right).$
This is still a substitution in which $x = 4 \cos u$ and
$\mathrm dx = 4\sin u\,\mathrm du$, so after substitution your integral becomes
$$
I = -16 \int \sqrt{\sin^2 u} \sin u\,\mathrm du.
$$
But the substitution $u = -\arccos\left(\frac14 x\right)$ implies that
$u \in [-\pi,0]$ and that $\sin u \leq 0.$ So you have
$\sqrt{\sin^2 u} = \lvert\sin u\rvert = -\sin u$ and
$$
I = 16 \int \sin^2 u \,\mathrm du.
$$
Continuing the evaluation from this point, you should eventually find that
$$
I = 8u - 8 \sin u \cos u + C.
$$
You now reverse the substitution using the rule $u = -\arccos\left(\frac14 x\right)$
so that $\cos u = \frac 14 x$ and
(recalling that $\sin u \leq 0$ for such a value of $u$)
$$
\sin u = - \sqrt{1 - \cos^u} = -\sqrt{1 - \frac1{16} x^2} = -\frac14 \sqrt{16 - x^2}.
$$
The result is
\begin{align}
I &= -8 \arccos\left(\frac14 x\right)
- 8\left(-\frac14 \sqrt{16 - x^2}\right)\left(\frac 14 x\right) + C \\
&= -8 \arccos\left(\frac x4 \right) + \frac12 x \sqrt{16 - x^2} + C ,
\end{align}
in agreement with the result that you get from the substitution that makes
$\sin u \geq 0.$
So you can have $\lvert\sin u\rvert = -\sin u$. But if at some point in the reverse substitution you forget that you are working with the case
$\lvert\sin u\rvert = -\sin u$, you may incorrectly replace $u$ with
$\arccos\left(\frac14 x\right)$ or replace $\sin u$ with $\frac14 \sqrt{16 - x^2}$,
or both, and then you will get an incorrect result (which you can confirm by differentiation).
Note that this answer is really just an example of the phenomenon described in a comment under the original question,
$\sqrt{16-x^2}=(-1)^k\sin u$ where $u \in [k\pi,(k+1)\pi].$
You could just as easily substitute
$u = 2\pi - \arccos\left(\frac14 x\right)$
and you would get the same result.
It should be easy enough to see that adding any multiple of $2\pi$ to the right side of that equation results in the same values of
$\sin u$ and $\cos u$ while causing the term that is linear in $u$
to add another constant to the constant $C.$
And we have already shown what happens for each choice of sign for
$\pm \arccos\left(\frac14 x\right).$
Also note that I have focused here on the evaluation of an antiderivative,
that is, a transformation from one function of a single variable to a family of functions of the same single variable.
I am not claiming that U-substitution in a definite integral,
$$
\int_{\phi(a)}^{\phi(b)} f(t) \mathrm dt
= \int_a^b f(\phi(u)) \frac{\mathrm d}{\mathrm du} \phi(u)\, \mathrm du
$$
(this version is from ProofWiki)
requires $\phi$ to be an invertible function.
What happens in the case of a definite integral is merely that when evaluating the integral on the right-hand side, at each individual point between $a$ and $b$ you need to evaluate the integral in a way that is consistent with the substitution at that point. This is explained in answers that various people (including myself) have written
here,
here,
here, and
here.
