TITLE: How does one visualize propositional logic?
QUESTION [9 upvotes]: I apologize if this is an under specified question, but I will try and provide some context. I am beginning to learn the foundations of propositional and predicate logic after starting a course in abstract algebra (groups, fields, rings, etc).
I find that predicate logic comes very easy to me because of its relation to set theory...the quantifiers make the whole visualization of "which elements make a statement true" very intuitive and easy to work through. However, I am struggling to find a similar feature with propositional logic (perhaps because  I have been trying to mimic my strategy from predicate logic). 
I understand that one can use truth tables to attack propositional logic...but the truth tables seem very...artificial and not particularly insightful. 
As an example, I commonly see this picture used to explain implication if $p$ then $q$ propositions:

Which, on its own, makes perfect sense. But I fail to see how this picture also represents $\neg p \lor q$, which is an equivalent form of $p \rightarrow q$. (As confirmed by truth tables)
If anything, the space "not $p$" would just be $q \land \neg p$...and therefore $\neg p \lor q$ would just be $(q \land \neg p) \lor q$...which is just $q$. But that means that $p \rightarrow q = q$...and that cannot be right, can it? That's sort of like modus ponens but missing the second step of decalring that $p$ is true. 
Regardless, I'm clearly not thinking about this subject in an effective way so if anyone can offer insight, I'd greatly appreciate it!
Edit: I am adding some pictures to clarify some confusion:
Answers offered below have made me realize that the picture previously provided is incomplete. There should exist another set, call it $E$ (for Everything) that contains, in addition to q, everything outside of q. This changes my understanding of what $\neg p$ means...but questions nonetheless remain.
$\neg p$ and $q$ are therefore illustrated as follows:
 
Therefore, $\neg p \lor q$ looks like this:

To me this means that every element in the universe satisfies the proposition of $\neg p \lor q$...which by equivalence, means that every element in the universe satisfies the proposition $p \rightarrow q$. Could I have further clarification?

REPLY [2 votes]: I find $A\implies B\space \equiv \space \neg (A \land \neg B)\space$ to be a more intuitive definition. 
If you want to visualize $A$ and $B$ as sets, you can use a Venn diagram that excludes only those elements of $A$ (left circle) that are not also elements of $B$ (right circle).

$x\in A \implies x\in B$
(Diagram from https://en.wikipedia.org/wiki/Material_conditional )
Note that all elements of $B$ satisfy this conditional relationshipship. So do all elements not in $A$.
You shouldn't depend too much on such diagrams, however. They are useful only for the most simple cases and won't be accepted as a valid proof in any advanced courses.