In a proof with multiple assumptions you have to choose one of them to be "blamed" for the contradiction.
Think to your example in terms of assumptions; you start with a couple of them (they can be two Lemmas already proved, or two hypotheses) :
$P→Q$ and $R→¬Q$.
Then we proceed "formally" as follows (I'll use the Natural Deduction proof system; for a good explanation of the rules to be used, see : Ian Chiswell & Wilfrid Hodges, Mathematical Logic (2007), Ch.2 : Informal natural deduction, page 5-on) :
1) $P$ --- assumed
2) $Q$ --- from 1) and $P→Q$ by $\rightarrow$-elim (modus ponens)
3) $R$ --- assumed
4) $\lnot Q$ --- from 3) and $R→¬Q$ by $\rightarrow$-elim (modus ponens)
5) $\bot$ --- from 2) and 4) by $\lnot$-elim [i.e. using the rule : "from $\varphi$ and $\lnot \varphi$, infer $\bot$]
6) $\lnot R$ --- from 3) and 5) by $\lnot$-intro [i.e. using the rule : "if from $\varphi$ we have derived $\bot$, then infer $\lnot \varphi$], "discharging" temporary assumption 3)
7) $P \rightarrow \lnot R$ --- from 1) and 6) by $\rightarrow$-intro, "discharging" temporary assumption 1).
Thus we have proved :
$P→Q, R→¬Q \vdash P \rightarrow \lnot R$.
As per the above answer, we can apply contraposition : $\varphi \rightarrow \lnot \psi \vdash \psi \rightarrow \lnot \varphi$ to conclude also :
$P→Q, R→¬Q \vdash R \rightarrow \lnot P$.
In the previous proof, we have chosen the assumptiom $R$ to be "blamed" for the contradiction. We can as well choose $P$.
If you rewrite it introducing $\lnot P$ in step 6) above, you will end exactly with : $R \rightarrow \lnot P$.
In order to "have a feeling" with the above application of logical rules, modify the above proof using a single assumption $P \land R$.
Due to the fact that :
$P \land R \vdash P$ and $P \land R \vdash R$ [by : $\land$-elim]
we can repeat the same steps until 5) : $\bot$.
In this case, we have only one assumption to be "blamed" : $P \land R$ and we conclude with :
$\lnot (P \land R)$.
This means that, in the presence of the two Lemmas or hypotheses : $P \rightarrow Q$ and $R \rightarrow \lnot Q$, we cannot "jointly assert" $P$ and $R$.
Thus, one of them must be "removed". Which one ? it's up to us ...