I understand that that by writing their logic tables they do not have the same truth values for corresponding inputs and because of this they are not equivalent.
But as an example consider the following statements. It is raining or it is sunny. If its raining, then it is not sunny. Why aren't these equivalent?
On
Your proposed equivalence would work if the 'or' is assumed to be an exclusive or, e.g. when we say that 'a whole number is even or odd' we do mean that 'if a whole number is even, then it is not odd'. Indeed, your 'sunny' vs 'raining' example suggests that maybe that is how you were thinking about the 'or', as it can't be both sunny and raining.
However, when logicians use the $\lor$, they mean the inclusive or, where it is possible for both disjuncts to be true. And, as such, just because one of the disjuncts is true does not mean that the other one has to be false.
On
But as an example consider the following statements. It is raining or it is sunny. If its raining, then it is not sunny. Why aren't these equivalent?
Because one does not entail the other, nor vice versa even.
Assume "It is rainy or it is sunny," holds. Further assume "It is not rainy". It would therefore be sunny. So we conclude "It is rainy or it is sunny" implies "If it is not rainy, then it is sunny."
Assume "If it is not rainy, then it is sunny" holds. It is either rainy, or it is not. In the first case, it is rainy. In the second case, because it is not rainy, therefore it is sunny. We conclude that "If it is not rainy, then it is sunny" implies that "It is rainy or it is sunny."
Therefore "It is rainy or it is sunny," is equivalent to "If it is not rainy, then it is sunny"
$$p\vee q \iff \neg p\to q$$
On the other hand.
Assume "If it is rainy, then it is not sunny" holds. It is either rainy, or it is not. In the first case, because it is rainy, therefore it is not sunny. In the second case, it is not rainy. We conclude "If it is rainy, then it is not sunny" implies that "It is not rainy or it is not sunny."
Assume "It is not rainy or it is not sunny" holds. Further assume "It is rainy". It would therefore be not sunny. So we conclude "It is not rainy or it is not sunny" implies "If it is rainy, then it is not sunny."
Therefore "If it is rainy, then it is not sunny" is equivalent to "It is not rainy or it is not sunny."
$$\neg p\vee \neg q \iff p\to\neg q$$
Furthermore $p\vee q \iff (p\wedge\neg q)\vee(p\wedge q)\vee(\neg p\wedge q)$
While $\qquad\neg p\vee\neg q \iff (p\wedge\neg q)\vee(\neg p\wedge\neg q)\vee(\neg p\wedge q)$
Because $p \lor q$ can be true either if $p \land \bar{q}$ or $\bar{p} \land q$ or $p \land q$.