Do I have to discharge an antecedent that I assume? For example, if I have the premise:
$P \rightarrow (Q \rightarrow R)$
Can I assume P to get:
$Q\rightarrow R$
And then assume Q to get R.
For reductio ad absurdum and arrow introduction I know that you have to discharge the assumptions that you use, I was just wondering if this is the case for assuming the antecedent.
Here is an argument that seems to require the assumption of the antecedents.
$E\rightarrow (\sim F \lor \sim (A \lor D)) \therefore E \land F \rightarrow \sim D \lor G$
 A: 
For example, if I have the premise:
$P \rightarrow (Q \rightarrow R)$
Can I assume P to get:
$Q\rightarrow R$
And then assume Q to get R.

You may do that, but discharging those assumptions gets you right back to $P\to(Q\to R)$
Suppose however that you assumed $Q$ then assumed $P$, to derive $R$. Discharging the assumptions will deduce $Q\to(P\to R)$ is entailed by the premise.
But $R$ is not entailed by $P\to(Q\to R)$, alone.
$R$ is entailed by $P\to (Q\to R)$, $P$, and $Q$.

Here is an argument that seems to require the assumption of the antecedents.
$E\to(\lnot F \lor \lnot (A \lor D)) \therefore E \land F \to \lnot D \lor G$

This is how a Conditional Proof works in Natural Deduction.   Here you assume $E\land F$, derive $\lnot D\lor G$, then discharge the assumption via conditional introduction.
$$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}\fitch{E\to (\neg F\lor\lnot (A\lor D))\quad:\textsf{Premise}}{\fitch{E\land F\quad:\textsf{Assumption}}{F\quad:\textsf{Conjunction Elimination}\\E\quad:\textsf{Conjunction Elimination}\\\lnot  F\lor\lnot(A\land D)\quad:\textsf{Conditional Elimination}\\\quad\vdots\\\lnot F\to \lnot D\quad:\textsf{Conditional Introduction}\\\quad\vdots\\\lnot(A\lor D)\to\lnot D\quad:\textsf{Conditional Introduction}\\\lnot D\quad:\textsf{Disjunction Elimination}\\\lnot D\lor G\quad:\textsf{Disjunction Introduction}}\\(E\land F)\to(\lnot D\lor G)\quad:\textsf{Conditional Introduction}}$$
A: You could do so, but the proof would be unfinished. You could discharge each premise in turn to get: 
$$P\implies [Q \implies R] \implies [P\implies [Q \implies R]]$$
If you introduced all of your premises at once, you could also prove:
$$[P\implies [Q \implies R] \land P \land Q] \implies  R$$
