Conditional Proof with Conditional within Antecedent

My first instinct is to assume $\ Q\implies\lnot R$ to construct a conditional proof, but would that require a subproof, in which I create another subproof where I assume the antecedent of the assumption (namely$\ Q $ ) to derive $\lnot R$ ?

No, not quite. You do assume the antecedent $Q \to \neg R$ to open a conditional proof, the conclusion of which will be the succedent $S \lor P$ -- but at the point where you've made this assumption, the implication $Q \to \neg R$ is already given (by that very assumption), so you do not need to open a subproof where you derive $\neg R$ from $Q$ yet again ($\neg R$ would not even be derivable from $Q$ without further assumptions). If you are to prove an implication the antecedent of which is yet another implication, you don't need to prove that implication. All you need to show is that if the antecedent implication holds, then the succedent follows. So instead, you work directly under that assumption $Q \to \neg R$ together with the two premises to derive the succedent of the implication, $S \lor P$.

I assume that the premises are there to ease the process of finalizing the proof, but as neither premise seem to directly ascertain any proposition, how would one use them to aid the proof, if all that is needed is to derive a conditional proof from an assumption of the antecedent?

You need both of the premises combined with the assumption $Q \to \neg R$ to derive the succedent $S \lor P$. The thing you will have under the subproof with assumption $Q \to \neg R$ will be a proof by cases, i.e., a disjunction elimination on the premise $Q \lor S$: You assume $Q$ and you assume $S$, derive the desired conlusion $S \lor P$ from both assumptions, and since by $Q \lor S$ you know that at least one of the assumptions must be true, you can be sure that the conclusion must be true, and get the implication succedent $S \lor P$ by disjunction elimination on the premise $Q \lor S$ and the two subproofs $Q \vdash S \lor P$, $S \vdash S \lor P$. This is the use of the second premise.

So what you need to do next is the first subproof required for he disjunction elimination, namely a subproof with assumption $Q$ and conclusion $S \lor P$. So you do need a subproof with assumption $Q$, but the goal of this subproof is not $\neg R$, but $S \lor P$. In that subsubproof, you have to do yet another subsubsubproof to derive $P$ from $\neg R$ with the help of $\neg P \to R$ (so this is where the first premise enters) by contraposition and double negation elimination: $\neg P \to R$ (premise) $\vdash \neg R \to \neg \neg P$ (contrapos.) $\vdash \neg R \to P$ (DNE). Once you've reached $P$, you can derive the desired succedent $S \lor P$ by a simple disjunction introduction.

For the second subproof required for the disjunction elimination, $S \vdash S \lor P$, you just need to do one more disjunction introduction on the assumption $S$ to get $S \lor P$.

Now that you've shown that the implication succedent $S \lor P$ follows from both cases $Q$ and $S$, you're ready to perform the disjunction elimination on the premise $Q \lor S$ and the two subproofs $Q \vdash S \lor P$, $S \vdash S \lor P$ to derive $S \lor P$.
All that happend in the subproof under the assumption $Q \to \neg R$, which serves as the antecedent of the assumption - and now that we have the succedent, we can finally do the conditional introduction on the subproof $Q \to \neg R \vdash S \lor P$ and arrive at the conclusion formula $(Q \to \neg R) \to (S \lor P)$.

Here is the full proof, typeset with this natural deduction proof editor and checker:

My first instinct is to assume $Q\to ¬R$ to construct a conditional proof, but would that require a subproof, in which I create another subproof where I assume the antecedent of the assumption (namely $Q$) to derive $¬R$?

You do not need to derive the assumption, you have assumed it. It is considered true within the context of the supproof raised by its assumption.

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{\neg P\to R\\Q\vee S}{\fitch{Q\to \neg R}{~\vdots~~\textsf{Somehow}\\S\vee P}\\(Q\to\neg R)\to(S\vee P)}$$

I assume that the premises are there to ease the process of finalizing the proof, but as neither premise seem to directly ascertain any proposition, how would one use them to aid the proof, if all that is needed is to derive a conditional proof from an assumption of the antecedent?

That is where assuming $Q$ is useful; as also is assuming $S$.   To eliminate the disjunction in the second premise, you next raise two sub-proofs, assuming $Q$ and $S$, aiming to derive $S\vee P$ in each (ie a proof by cases).

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{\neg P\to R\\Q\vee S}{\fitch{Q\to \neg R}{\fitch {Q}{~\vdots~~\textsf{Somehow}\\S\vee P}\\\fitch{S}{S\vee P}\\S\vee P}\\(Q\to\neg R)\to(S\vee P)}$$

but as neither premise seem to directly ascertain any proposition, how would one use them to aid the proof,

Do it indirectly, with a sub proof showing that $\neg R$ entails $(P\vee S)$ under the premises. (Or if allowed, use Modus Tollens: $\neg R, \neg P\to R\vdash \neg\neg P$; although the subproof is easy enough.)   This is where the second premise is utilised in this proof.

$$\fitch{\neg P\to R\\Q\vee S}{\fitch{Q\to \neg R}{\fitch {Q}{\neg R\\\fitch{\neg P}{R\\\bot}\\\neg\neg P\\P\\S\vee P}\\\fitch{S}{S\vee P}\\S\vee P}\\(Q\to\neg R)\to(S\vee P)}$$

how would one use them to aid the proof, if all that is needed is to derive a conditional proof from an assumption of the antecedent?

Assuming the antecedent is often not all that is needed to derive the consequence.