From my understanding,
Proof by contrapositive: Prove $P \implies Q$, by proving that $\neg Q \implies \neg P$ since they are logically equivalent.
Proof by contradiction: Prove $P \implies Q$ by showing that $P \wedge \neg Q$ yields an absurdity and hence false. So $\neg (P \wedge \neg Q)$ is equivalent to $\neg (\neg (P \implies Q))$ and $P \implies Q$ by double negation so showing that $\neg (P \wedge \neg Q)$ proves $P \implies Q$.
If the absurdity derived during the procedure for a proof by contradiction is $P \wedge \neg Q \implies\neg P$, we have essentially already proven $P \implies Q$ by contrapositive since $\neg Q \implies \neg P$ is precisely the required condition for proof by contrapositive. But $(P \wedge \neg Q) \implies \neg P$ is also a contradictory statement which means that $P \implies Q$ must be true.
Now the question is this. Is this proof by contradiction still a valid form of proof even though its a proof by contrapositive in disguise? To me, this proof by contradiction also seems to be a valid proof as it does seem to satisfy the conditions(if they are correct) for proof by contradiction.
Additionally, if you have a contrapositive proof, so you have shown that $\neg Q \implies \neg P$, is it possible to rephrase this in a proof by contradiction by supposing that $P \wedge \neg Q$ instead of just $\neg Q$.
If this is the case, what is the point in distinguishing proof by contradiction from proof by contrapositive?
edit: My thought is that proof by contrapositive is a direct proof while proof by contradiction, in this case, depends on the validity of the double negation law which apparently isn't valid in intuitionistic logic.
Yes it is valid... it doesn't really matter if it's something else 'in disguise', just that is it correct. And deriving $\lnot P$ from $P\land\lnot Q$ is certainly leads to a contradiction that implies $\lnot (P\land \lnot Q)$ is true, which implies that $P\to Q$ is true. One thing to note (that I think you have noticed based on your second question) is that you have an additional assumption of $P$ open and available for use when you derive $\lnot P,$ unlike in the case of just deriving $\lnot Q\to \lnot P$ by assuming $\lnot Q$ and deriving $\lnot P.$
So the second question amounts to whether it is admissible to assume $P$ in a proof of $\lnot Q\to \lnot P.$ It is, and the easiest way to see this is reasoning semantically and using the completeness/soundness theorem. If $P\vdash \lnot Q\to \lnot P$ then every interpretation in which $P$ is true has $Q$ true, which means precisely the same thing as $\vdash P\to Q,$ so we have $\vdash \lnot Q\to \lnot P.$
As a result, proof by contrapositive is essentially a special case of what you're calling proof by contradiction, where the contradiction takes the special form $\lnot P$ contradicting with $P.$
I think the reason you might have seen framing contrapositive proofs as proofs by contradiction discouraged is for style reasons. Unless the outstanding assumption of $P$ is used (unnecessarily), the part of the proof outside the inner proof of $\lnot Q\to \lnot P$ is just boilerplate that can be omitted. It is also more informative to call it a proof by contrapositive since that is a special case of contradiction.
(As Derek indicates in the comments, proof by contrapositive is not intuitionistically valid, so it doesn't have anything to do with this.)