I am reading the Pu's lemma on Wiki, I don't know the red line. Seemly, the author omit the detail of red line. Therefore, I found the Pu's paper, But seemly, Pu also omit it, as picture bottom. And the reference $[2]$ of Pu is not in English. The 188 of $[3]$ seemly be not useful. I think the useful part is 185, but it is a little complex for me. Who can explain the average over a group in a simple way ? And why $f_{\text{new}}$ is constant ?
From the Wiki of Pu's Lemma
From the 60th page of Pu's paper
A compact Lie Group is a compact smooth manifold, right? And it also has a left-G-invariant measure (the Haar measure --- this may be the idea that you're missing). And by computing the measure of the whole manifold $G$, you can normalize this total measure to $1$.
Now if you have a real-valued function $f$ defined on $G$, you can integrate $f$ with respect to this measure, and the resulting number will be "the average of $f$ over $G$.
The Wikipedia page for Haar measure points out that a left-invariant measure on a Lie Group was known before the Haar measure was, but I honestly don't recall the history there.
As an example, if you parameterize the unit circle $S^1$ with the interval $[0, 2\pi]$ in the usual way, $t \mapsto (\cos(t), \sin(t))$ and use $dt$ as your measure, you find that the total measure of $S^1$ is $2\pi$. So you can replace $dt$ with $\frac{1}{2\pi} dt$ to get a normalized measure.
Now let's suppose that your function (defined in "t" coordinates) is, say $$ f(t) = \cos^2(t) $$ which is to say, the square of the $x$-coordinate of a point on the circle. To find its average value, we compute \begin{align} \int_{S^1} \cos^2(t) \frac{1}{2\pi} dt &= \int_0^{2\pi} \cos^2(t) \frac{1}{2\pi} dt \\ &= \frac{1}{2\pi} \int_0^{2\pi} \frac{1 + \cos(2x)}{2} dt \\ &= \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{2} dt + \frac{1}{2\pi} \int_0^{2\pi} \frac{\cos(2x)}{2} dt \\ \end{align} The second integral is zero, and the first evaluates to $\pi$, so the final answer is $\frac{1}{2}$.