For the statement A implies B, write the statement NOT B implies NOT A.
The logical form of the implication you're given is $a > 0 \implies\neg \exists M \alpha$, where $\alpha$ just stands for the second inequality that you wrote in your question. The contrapositive is $\exists M \alpha \implies a \leq 0$.
The main thing to recognize is that the consequent of the given conditional is a negated existential sentence: "There does not exist ..." In the contrapositive, this sentence becomes an ordinary existential sentence: "There exists..."
Added: You seem to be confusing quantifier equivalences and contrapositives. It's true that $\neg \exists \alpha \equiv \forall \neg \alpha$ and dually $\neg \forall \alpha \equiv \exists \neg \alpha$. But we needn't use those equivalences here.
We use the contrapositive equivalence $A \implies B \equiv \neg B \implies \neg A.$
Now, again, we are given $a > 0 \implies \neg \exists M \alpha$. By the contrapositive equivalence this becomes $\neg \neg \exists M \alpha \implies \neg(a > 0)$. We can remove the double negation, and use $\neg(a>0) \equiv a \leq 0$ to get what I wrote above (which agrees with the textbook answer).
The thing to notice is that, in the contrapositive sentence, there is no longer any negation to "push in" over a quantifier, so the quantifier equivalences that I wrote above do not apply.
To focus on the part that's confusing you, I suggest thinking about the latter part as a statement about the number $M$: specifically, let's declare $P_{a,b,c}(M)$ to be the statement
"$ax^2+bx+c\leq M$ for every real number $x$."
So the original statement is: "If $a>0$, then there does not exist any $M$ for which $P_{a,b,c}(M)$ is true."
Then the contrapositive statement is: "If there does exist an $M$ for which $P_{a,b,c}(M)$ is true, then $a\leq 0$".
Let $a, b, \text{and } c$ be real numbers. If $a>0$, then there does not exist a real number $M$ such that, for every real number $x$, $ax^2+bx+c≤M$.
Let's break this down. The first part just tells us that we are working with real numbers. This will be irrelevant to the discussion as those are the only kind of numbers we are working with, so any logical manipulation will have to be with real numbers.
The "if" part of the statement: $A = a>0$. So right away we know the negation of this is $a \leq 0$.
Now, the "then" part. The claim is that for every real number $x$, there does not exist a real number $M$ such that... In other words, there is no single $M$ such that ... for every real number $x$. Or even more simply and ambiguously, there is no "one-size-fits-all" $M$ for all the real numbers. And the condition is that $ax^2+bx+c≤M$. Just to summarize the "then" part has this format: "No single $M$ for all the $x$'s such that this happens: $ax^2+bx+c≤M$." Summarizing it this way makes it easy to come up with the negation. I just have to come up with a "generic" counter-example.
The negation to the "then" part is: There is a single $M$ for all the $x$'s such that this happens: $ax^2+bx+c≤M$. Let this statement be called $\neg B = C$.
Now you just put it together: (If $\neg B, \text{then } \neg A) \iff (C \implies \neg A)$.