The Chaz alluded, in a comment, that by sufficiency of "or" one can reason that both implications are true.
Note that $A \rightarrow B$ is equivalent to $\lnot A \lor B$ (you can confirm this using a truth table for each and convince yourself that they yield the same "output" - truth value - for any given combination of truth values of A and B.
Now, this is what always helped me when first learning logic:
Let the truth value of $A$ be false, of $B$, true. Then $A \implies B$ is an example of case 1 in which we have a false statement implying a true statement. Now, if we take the equivalence of $A \implies B$ and $\lnot A \lor B$, then in this case, $\lnot A$ is $\lnot\text{False}$, i.e. $\lnot A$ is a true statement. Since $\lnot A$ is a true statement, then so is $\lnot A \lor \text{True}$, and so is $\lnot A \lor \text{False}$ (if $B$ happened to be false), since a disjunctive ("or") statement evaluates to true whenever (at least) one of its disjuncts is true. Since $\lnot A$ is true, then $\lnot A \lor X$ is true no matter what the truth value of the statement $X$ happens to be.
A more "non-mathematician's" explanation might use a down-to-earth, not far-fetched example, like the following. Suppose the forecast predicts a 50/50 chance of rain tomorrow. I immediately assert that "If it rains tomorrow, then I'll take an umbrella with me to campus." Okay. So, now it's tomorrow:
scenario 1. It isn't raining, (and it doesn't rain at all), but I took my umbrella with me in case of rain. False --> True.
scenario 2. It isn't raining (and it doesn't rain at all). Good for me because I decided not to take my umbrella (since I have enough to lug around with me). False --> False.
Does that help?