I agree that the link you provide to the proof offered by Proof Wiki is "self-reliant", and seems, to me at least, to invoke the very rule in which we are interested. In a sense, it can be seen as a "model" proof exemplifying the general case of invocation of Disjunction Elimination, $\lor \mathcal E$.
So, yes, this can sort of be thought of as an "axiom": a basic rule of inference we take to be valid, and from which other rules of inference may be derived. It does rely on modus ponens, twice. But the principle that, if, given the truth of $p \lor q$, and some conclusion $r$ follows from $p$, and the same conclusion $r$ follows from $q$, then we can conclude that $r$ is true.
It might be reassuring to note that the argument characterizing "Disjunction Elimination" can be expressed as follows:
$$(((P \to Q) \land (R \to Q)) \land (P \lor R)) \to Q\tag{Tautology}$$
which proves to be tautologically (necessarily) true $(\dagger)$, and so we have very good reason to accept the rule of inference as valid.
$(\dagger)$: Exercise - Confirm, using a truth-table, that the given tautology is, in fact, tautologically true.
Best Answer
Not really. After all, you shouldn't expect to be able to derive components from an 'or' statement: after all 'TRUE or FALSE' is true, but you can only possibly derive 'TRUE' as a true statement. You can't possibly prove 'FALSE'.
This rule is, indeed, one of the trickiest to understand. This is because it doesn't just take statements as its inputs. Instead, two of the inputs - the large vertical boxes in the rule - are entire proofs. If you have '$p \vee q$' and a proof that assuming $p$ you can conclude $\chi$ and a proof that assuming $q$ you can conclude $\chi$... then from all three of these things - the disjunction and the two proofs - you can conclude $\chi$.
This rule might be easier to appreciate if you look at it from the viewpoint of how we might naturally prove things, where it looks exactly like a proof by cases:
The disjunction elimination rule is just the formalisation of this sort of argument.