Direct Proof with more than one conditional. If I have a statement such as $p \implies (q \implies r)$ that I want to prove it with a direct proof. Would I assume $p \land q$ are true and then follow the axioms and previously proven theorems to reach $r$? Does that mean that the statement only holds when $p \land q$ are true?
 A: Here are the questions:

If I have a statement such as $p⟹(q⟹r)$ that I want to prove with a direct proof, would I assume $p ∧ q$ are true and then follow the axioms and previously proven theorems to reach $r$?

Suppose one already knows that $r$ is true. Then one would prove this conditional by assuming $p$ and $q$. One does not need to assume them conjoined as $p ∧ q$. 
Here is an example of such a proof in a Fitch-style proof checker:


Does that mean that the statement only holds when $p ∧ q$ are true?

Consider the truth table for the conditional:

Note that the only time the conditional is false is when $p$ is true, $q$ is true, but $r$ is false. For all other valuations, the conditional is true. So the statement doesn't only hold when $p$ and $q$ are true. In fact there is one valuation where the conditional is false when $p$ and $q$ are both true.
It may be better to say that the statement holds when $r$ is true. This is why I assumed $r$ as a premise in the proof.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html
A: For the first part of the question your are correct.
For the second:

Does that mean that the statement only holds when $p∧q$ are true?

No, if $p$ is false then $p\Rightarrow(q\Rightarrow r)$ also holds because false implies anything. 
Similarly if $q$ is false, then $q\Rightarrow r$ is true and thus $p\Rightarrow(q\Rightarrow r)$ also holds (anything implies the truth). 
A: 
If I have a statement such as $p⟹(q⟹r)$
  that I want to prove it with a direct proof. Would I assume $p∧q$ are true and then follow the axioms and previously proven theorems to reach $r$?

To be clear: You assume $p$ and you assume $q$, with the hope of deriving $r$ by some valid inferences.   Should you be able to so derive $r$ under the two assumptions, you may then discharge them, one by one (using conditional introduction.$$\begin{array}{|l}\ddots\\\hline\quad \begin{array}{|l}p\\\hline\quad\begin{array}{|l}q\\\hline\vdots \\r\end{array}\\q\to r\end{array}\\p\to (q\to r)\end{array}$$

Does that mean that the statement only holds when $p∧q$
  are true?

No.   You are claiming that $p\to (q\to r)$ holds, that is, validly derived from whatever premises you have (which to not need to establish valuations for either $p$ or $q$).
Indeed the statement may be derived under the premise that $\lnot q$, vai ex falso quodlibet.
$$\begin{array}{|l}\lnot q\\\hline\quad \begin{array}{|l}p\\\hline\quad\begin{array}{|l}q\\\hline\bot \\r\end{array}\\q\to r\end{array}\\p\to (q\to r)\end{array}$$
