$∃$ quadrilaterals $x$ such that if $x$ is a parallelogram, then $x$ is a kite.
I understand the above statement to be false.
Using the formulae $$¬(∃x∈U)[P(x)]≡(∀x∈U)[¬P(x)]\\ ¬(p⟹q)≡¬(¬p∨q)≡¬¬p∧¬q≡p∧¬q,$$ its negation is:
$∀$ quadrilaterals $x, x$ is a parallelogram and $x$ is not a kite.
Since the original statement is false, the negation of that statement should be true. However, the above negation is also false! I am not sure what I'm doing wrong.
I think the best way to learn how to work with statements involving quantifiers and implications is to write out what they mean in words
The first statement says
There is a quadrilateral about which you can say that if it's a parallelogram then it's a kite.
That statement is true, because there are quadrilaterals that are not parallelograms. Take one of those irregular quadrilaterals for your $x$. Then the implication
If $x$ is a parallelogram then it's a kite.
is true for that particular $x$ since they hypothesis is false. (That's often confusing for students at first.)
$∃$ quadrilaterals $x$ such that if $x$ is a parallelogram, then $x$ is a kite.
I understand the above statement to be false.
This is your only mistake. It is understandable, because an existentially-quantified conditional $$\exists x\,(Px\to Kx)\tag1$$ is unintuitive to process, because it sounds unnatural and we are subconsciously tempted to read it as $$\forall x\,(Px\to Kx)\tag2$$ instead. Statement $(2)$ is false; on the other hand, putting $x=\text{kite}$ shows that the given statement, statement $(1),$ is true; that is, there indeed exists a quadrilateral, such as a kite, such that if it's a parallelogram then it's a kite (a kite is not a parallelogram, and every conditional with a false hypothesis is vacuously true).
