In section 8.1 of Brown Rust Book, there is the following question:

Determine whether the program will pass the compiler.
If it passes, write the expected output of the program if it were executed.

fn main() {
  let mut v: Vec<i32> = vec![1, 2, 3];
  let mut v2: Vec<&mut i32> = Vec::new();
  for i in &mut v {
    v2.push(i);
  }
  *v2[0] = 5;

  let a = *v2[0];
  let b = v[0];
  println!("{a} {b}");
}

I thought that this obviously doesn't compile because we are taking multiple mutable references to v. But it turns out I was wrong. The program compiles fine.

Which is weird because in section 4.3 of the same book, they say the following:

Rust’s borrow checker does not contain different places for a[0], a[1], and so on. It uses a single place a[_] that represents all indexes of a.

They were talking about arrays in that section, but the same applies to vectors.

For example, this doesn't compile:

fn main() {
    let mut v = vec![0, 1, 2, 3];
    let x = &mut v[1];
    let y = &mut v[2];
    *x += *y;
    println!("{v:?}");
}

Output:

error[E0499]: cannot borrow `v` as mutable more than once at a time
 --> src/main.rs:4:18
  |
3 |     let x = &mut v[1];
  |                  - first mutable borrow occurs here
4 |     let y = &mut v[2];
  |                  ^ second mutable borrow occurs here
5 |     *x += *y;
  |     -------- first borrow later used here
  |
  = help: use `.split_at_mut(position)` to obtain two mutable non-overlapping sub-slices

For more information about this error, try `rustc --explain E0499`.

But this totally compiles:

fn main() {
    let mut v = vec![0, 1, 2, 3];
    let mut iter = v.iter_mut();
    let x = iter.next().unwrap();
    let y = iter.next().unwrap();
    *x += *y;
    *y += *x;
    println!("{v:?}");
}

Output:

[1, 2, 2, 3]

So my question is:
How does this work?
Is the iterator doing something similar to split_at_mut for each element under the hood?

Disclaimer:
I'm a Rust beginner so sorry if the question is stupid or was answered before.

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