Can $P \implies Q$ be represented by $P \vee \lnot Q $? Source: p 46, How to Prove It by Daniel Velleman
Though the author writes $Q$ (the original apodosis) as 'You'll fail the course',
for brevity I shorten $Q$ to 'You fail'.

Let $P$ be the statement “You will
  neglect your homework” and $Q$ be “You fail.”
  Then “You won’t neglect your homework, or you fail.” $ \quad = \color{green}{\quad \lnot P \vee Q}$.
But what message is the teacher trying to convey with this statement?
  Clearly the intended message is
  “If you neglect your homework, then you fail,” or in other words 
  $P \rightarrow Q.$
  Thus, in this example, the statements $\lnot P \vee Q$ and $ P \rightarrow Q $ seem to
  mean the same thing.

Why cannot the bolded be symbolised as
$\color{darkred}{P \vee \lnot Q}$ = "You neglect your homework, or you fail not." ?
I am trying to intuit Material Implication: intuitively, how does $P \Longrightarrow Q \equiv \color{green}{\lnot P \vee Q}$ ?

This follows simply from the Law of Excluded Middle: $P \lor \neg P$ for all propositions $P$.

Let's assume $P \rightarrow Q$ and deduce $Q \lor \neg P$. 
By the Law of Excluded Middle, we have either $P$ or we have $\neg P$. We do a case analysis over which one is true:
If we got a $P$, by our assumption, we can deduce $Q$. 
If we got a $\neg P$, well... we stick with $\neg P$.
Having covered both cases, we know one or the other is true, so we have $Q \lor \neg P$.

Let's go the other way now, to show the statements are logically equivalent. We'll assume $Q \lor \neg P$ and deduce $P \rightarrow Q$.
We assume we have a $P$, and our goal is to produce a $Q$, and this will prove $P \rightarrow Q$.
Do case analysis on our assumption $Q \lor \neg P$. 
If we have $Q$, we've have fulfilled our obligation.
If we have a $\neg P$, on the other hand, we notice that we also have a $P$ just hanging around in our current list of assumptions. This gives us a contradiction, and by Principle of Explosion, we may conclude $Q$. This again fulfills our obligation.
