No, the `return` is not ignored; it just has special semantics in
generator functions; see <https://3v4l.org/rCAX4> and
<https://github.com/php/php-langspec/blob/master/spec/14-classes.md#class-generator>.

But the 2nd example does not call yield at all - if (false) {} pattern is used a lot to comment code to keep it refactorable - why is this allowed and where is this behaviour defined?

Any function containing yield is a generator and follows generator semantics. Whether the yield actually gets executed does not matter.

Please, this is generators 101.

This should then be - like return - not allowed, shouldn't?

$u = fn() => yield 5;
echo $u();

Why would that be disallowed? It's equivalent to

function() {
    return yield 5;
}

which is not terribly useful, but perfectly legal code. It yields one value, accepts one value and specifies the accepted value as the coroutine return value.

Ok, just example how to mix non-generator and generator return.

class Cl {
    public function __get($name) {
        if ($name === 'gen') {
            return (fn() => yield 'a')(); // return generator
        } else {
            return ['b']; // return array
        }
    }
}

$cl = new Cl();
var_dump($cl->gen);
var_dump(iterator_to_array($cl->gen));
var_dump($cl->not_gen);

In this case, Cl::__get() is a normal function, not a generator
function.  The anonymous function inside of Cl::__get() is a
generator function, though.