Your understanding is correct, but it may help to address a few other things you seem to be dealing with. Here is your problem as it currently stands:
Problem: Suppose $P \to (Q \to R)$. Prove that $\lnot R \to (P \to \lnot Q)$ using truth tables.
It may be helpful to rephrase this symbolically; you are being asked to show that the following implication is always true:
Rephrased problem: $[P \to (Q \to R)]\to[\lnot R \to (P \to \lnot Q)]$
For convenience, I'll use the following symbolism:
- $\Omega: P\to(Q\to R)$
- $\Phi : \neg R\to(P\to\neg Q)$
The cool thing is that you have noticed, via truth table, that the truth values of $\Omega$ and $\Phi$ are all the same. Thus, you can see that $\Omega\to\Phi$, as desired, but you can also see another important thing: $\Phi\to\Omega$. There's a special term for when this happens: equivalence. By your own observations, you can see that $\Omega\to\Phi$ and $\Phi\to\Omega$, a conjunction of two implications in the form $p\to q$ and $q\to p$. A statement of this form is called an equivalence and is often denoted by $p\Leftrightarrow q$; you can also express the equivalence of $p$ and $q$ by writing $p\equiv q$, where the symbol "$\equiv$" denotes equivalence and this is reflected in how that symbol is actually typeset (i.e., \equiv).
In your own problem, you have that $\Omega\Leftrightarrow\Phi$ (or simply $\Omega\equiv\Phi$); that is, $\Omega$ and $\Phi$ are equivalent. And you can actually see this by looking at your truth table; construct a column for $(\Omega\to\Phi)\land(\Phi\to\Omega)$. The entire column will have only truth values. This means that $\Omega\Leftrightarrow\Phi$ is what is called a tautology, a compound statement that is true for all truth values of the individual statements.
Hopefully that rather long-winded explanation sheds some light on the matter. As we have just established via truth table, $\Omega\equiv\Phi$. If you were asked to prove that $\Omega$ and $\Phi$ are equivalent though, I would encourage you to not use a truth table unless you absolutely have to. I'll outline a little proof below that shows $\Omega\equiv\Phi$, and hopefully this will show you the advantage(s) of not going the truth table route. Feel free to comment if a step does not make sense.
Problem: Show that $P \to (Q \to R)$ and $\lnot R \to (P \to \lnot Q)$ are logically equivalent.
Proof. First note the following for statements $p,q,r$:
$$
p\to q\equiv\neg p\lor q\qquad\text{and}\qquad \underbrace{p\lor(q\lor r)\equiv(p\lor q)\lor r}_{\text{associativity of $\lor$}}
$$
If you have not encountered the above equivalences, simply construct truth tables to see how they are equivalent. They will be used in the proof below though:
\begin{align}
P\to(Q\to R)&\equiv\neg P\lor(\neg Q\lor R)\tag{since $p\to q\equiv\neg p\lor q$}\\[0.5em]
&\equiv R\lor(\neg P\lor\neg Q)\tag{by assoc. of $\lor$}\\[0.5em]
&\equiv \neg R\to(P\to\neg Q).\tag{since $p\to q\equiv\neg p\lor q$}
\end{align}
This concludes the proof; notice how much easier/faster that was than constructing a truth table.
