Halting problem vs Universal Language
So first Q1:
At first the definitions of universal language and the universal machine.
The universal language is defined as you mentioned in the question:
$$ U= \{(M,w)\:|\:\text( TM\:M\:accepts\:w)\} $$
The Universal Turing Machine on the other hand is defined as follows:
A machine, which can simulate an arbitrary TM on arbitrary input. Or you can say: It's a TM U, which you can give a random other TM M and an input w. The Universal TM then simulates the given TM M on that input w.
So the Universal Turing Machine accepts, if and only if the TM M accepts w (because it simulates M on w). It rejects, if and only if, the TM M rejects w. It halts, if and only if the TM M is halting on w (because it simulates M on w). So the Universal Turing Machine accepts the above defined Universal Language.
Now to the question: The definition of the link is a little bit misleading. The Universal TM doesn't accept HALT on default, but you can modify the Universal Turing Machine in a simple way to accept HALT: The reject states are replaced by accept states. The now modified UTM (let's call it U') is accepting HALT. Because now it always accepts, if the machine M is accepting or rejecting the input w, without getting stuck in a loop: U' is accepting, if M is halting on w.
Then Q2:
A &quot;TM M is accepting a language L&quot; means, that it definitely halts on w $\in$ L, but does not mean it halts on w $\notin$ L (it can loop forever). So that means, if the U' is accepting HALT, it holds for sure, only if $(M,w) \in HALT$. If $(M,w) \notin HALT$, the universal TM can halt or can loop forever. So it isn't deciding HALT, it's just accepting HALT.
(Actually it's the slightly modified Universal TM, which is accepting HALT as mentioned in answer Q1.)
Q3:
If you mean the Universal Language with L(UTM) and the halting language with L(HTM), then yes L(UTM) $\neq$ L(HTM).
Q4:
First three points you made are correct.
In the last point you have to be more exact: The (modified) UTM can accept the Halting Language, but cannot decide it. So you can check, whether M halts on w, but it may loop forever.