When using a bounded existential quantifier, why can't I use an implication instead of a conjunction when expanding it to an unbounded quantifier? 
  ∃
  x
  ∈
  A
  P
  (
  x
  )
  ∨
  ∃
  x
  ∈
  B
  P
  (
  x
  )




=
∃
x
(
x
∈
A
∧
P
(
x
)
)
∨
∃
x
(
x
∈
B
∧
P
(
x
)
)


  (
  ∀
  x
  ∈
  A
  P
  (
  x
  )
  )
  ∧
  (
  ∀
  x
  ∈
  B
  P
  (
  x
  )
  )

  
    
  
  =
  (
  ∀
  x
  (
  x
  ∈
  A
  →
  P
  (
  x
  )
  )
  )
  ∧
  (
  ∀
  x
  (
  x
  ∈
  B
  →
  P
  (
  x
  )
  )
  )

Why does the first expansion turn the expression inside the parentheses to a conjunction, while the second turns the expression into an implication statement?
When should I turn a bounded quantifier to a conjunction of the bound or an implication of the bound?
Does it have to do with the Universal vs existential bound?
I understand why the universal bound expansion is the implication, but I don't understand why the existential qualifier expansion cannot have the implication statement instead of the conjunction.
 A: Does

There is some $A$ which has property $P$

mean

There is something which is in $A$ and has property $P$

or

There is something which, if it's in $A$, has property $P$?

These statements correspond to "$\exists x\in A(P(x))$," "$\exists x(x\in A\wedge P(x)),$" and "$\exists x(x\in A\rightarrow P(x))$," respectively.
Intuitively it should be clear that the third candidate has a "silly" way of being true: as long as there is a single object which is not in $A$ in the first place, the statement "$\exists x(x\in A\rightarrow P(x))$" is true by virtue of the behavior of the material conditional. In particular, "$\exists x(x\in A\rightarrow P(x))$" is automatically true if $A$ is empty (and the universe isn't). This is pretty obviously not the situation described by the first statement! Note that this is clearer if we think explicitly in terms of "truth conditions" (or "falsity conditions") rather than trying to just "guess the meaning" of the statement in question. Complicated logical expressions are a bit unintuitive, and it's often helpful to think of them a bit abstractly.

This leaves the question of why we don't see a "simple duality" between the universal and existential versions. In fact there is a duality, but it's obscured a bit by the notation.
The key duality between $\forall$ and $\exists$ is given via negation: we have $$\forall x\varphi(x)\iff\neg\exists x\neg\varphi(x)\quad\mbox{and}\quad\exists x\varphi(x)\iff\neg\forall x\neg\varphi(x).$$ A duality of exactly the same form is at play with the bounded quantifiers: "$\forall x\in A$" is the same as "$\neg \exists x\in A\neg$," and similarly "$\exists x\in A$" is the same as "$\neg\forall x\in A\neg$." This should make sense when thinking in terms of natural language ("Something in $A$ has property $P$" is the same as "It is not the case that everything in $A$ has property $\neg P$"), and granting this duality we can justify one symbolization of the bounded quantifier in terms of the other.
For example, you're happy with the rephrasing of "$\forall x\in A(P(x))$" as "$\forall x(x\in A\rightarrow P(x))$," and less happy with the rephrasing of "$\exists x\in A(P(x))$" as "$\exists x(x\in A\wedge P(x))$." So let's convince ourselves of this latter equivalence, using the former better-understood equivalence and the above-mentioned duality between the bounded quantifiers. I've broken things into separate hints since I think that's ultimately more useful; it starts, as all good things do, with a (double) double negation:

 By applying double negation elimination twice, once to the "outside" and once to the "inside," the original statement $\exists x\in A(P(x))$ is equivalent to $$\neg(\color{red}{\neg\exists x\in A(\neg}\neg P(x))).$$ This looks messier, but that highlighted bit is quite nice since we can ....


  ... apply the bounded quantifier duality described above! We know that the red "$\neg\exists x\in A\neg$" can be replaced with "$\forall x\in A$," so the above is in fact equivalent to $$\neg\color{red}{(\forall x\in A(\neg P(x)))}.$$ The highlighted bit above can be simplified because it ...


 ... is in terms of a bounded quantifier we already understand! So we know how to unpack it, and we get $$ \color{red}{\neg\forall x}(x\in A\rightarrow\neg P(x)).$$ That bit in red reminds us of ...


 ... the way negation plays with ordinary quantifiers! We now use the original duality between the universal and existential quantifiers to bring that outermost negation inside a quantifier: "$\neg\forall$" is the same as "$\exists\neg$," and so we get $$\exists x(\color{red}{\neg(x\in A\rightarrow\neg P(x))}).$$

After negating the conditional and killing off one last double negation, this gives us, as desired, $$\exists x(x\in A\wedge P(x)).$$
I think the key point, in terms of convincing yourself that there really is a tight parallel between the universal and existential bounded quantifiers, is the result of the fourth step above, that $$\exists x\in A(P(x))\mbox{ means }\exists x(\color{red}{\neg}(x\in A\rightarrow\color{red}{\neg} P(x))).$$ This nicely parallels a fact you already understand, namely that $$\forall x\in A(P(x))\mbox{ means }\forall x(x\in A\rightarrow P(x)),$$ the difference being a careful insertion of two negation signs. It's just that in rewriting the bounded existential quantifier we keep going ... and, while we wind up with something easier to use in the end, this obscures the simple parallel between the two types of bounded quantifier.
