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.
I would say that all three forms are equivalent. They all express your assumptions followed by the consequence of those assumptions. The third one would be more easily translated into a purely symbolic representation, but it's worth noting that mathematics was a rhetorical art long before symbolism. With that in mind, they all state the same idea, that a certain property follows inevitably from certain other properties, and the rest is just stylistic.
You can see some guides for the difference between "suppose" and "let" here. It boils down to "Let $x$ be something" means you're telling me that you're using $x$ as a shorthand for an object with a certain property, while "suppose" can be used for the same thing and also for pretending something is true, such as "Suppose this theorem is true, here are the consequences". It wouldn't make much sense to say, "Let this theorem be true," because you can't just declare a theorem to be true the same way you can claim your arbitrary label $x$ represents a object with a certain property.
It's all very subtle, I hadn't thought about it before with absolute rigor. I doubt most people do. The difference is almost colloquial to math writing, a linguistic quirk rather than a mathematical one.
About finding the contrapositive, all the statements are the same, so we can choose one arbitrarily. I'll pick the third formulation for clarity. If we start with "If $H$ and $A$, then $B$", the contrapositive is clearly "If $\neg B$, then $\neg H$ or $\neg A$", with the antecedent being flipped around by DeMorgan's law. Also note that I combined all the $H_i$'s into one $H$ since it amounts to the same thing.
Now we're trying to prove an OR statement. You could prove it by cases, but there's another way. The statement "If $P$, then $Q$" is equivalent to the statement "$\neg P$ or $Q$". So likewise, if we have "$P$ or $Q$", we equivalently prove "If $\neg P$, then $Q$". It's like proving the OR statement by saying, "Ok, first assume $P$ is true. So the OR statement is true and we're done. Now assume $P$ is not true. We have to show $Q$ is true or else the OR statement isn't always true."
Applying it back to our example, we can prove "If $H$, then $\neg A$" instead of the OR statement in the consequent. So now we're saying, "If $\neg B$, then if $H$, then $\neg$A". That's the same as "If $\neg B$ and $H$, then $\neg A$", just use the earlier trick to rearrange it. So now you can rephrase it as "Suppose $H$. If $\neg B$ then $\neg A$" since those formulations are the same.
So you can keep it in the same domain, it all amounts to the same thing. It all depends on what you want to emphasize.
Best Answer
The negation of “If it is raining, then I will bring my umbrella” is not “If is not raining, then I will not bring my umbrella.” The latter is equivalent to the converse of the former.
In symbols, $$ \neg(p \rightarrow q) = \neg((\neg p) \lor q) = p \land \neg q $$ while $$ (\neg p) \rightarrow (\neg q) = (\neg(\neg p)) \lor (\neg q) = (\neg q) \lor p = q \rightarrow p $$
Back into words, the negation of “If it is raining, then I will bring my umbrella” is “It is raining and I won't bring my umbrella.”