First, I know that you know that the inference rules aren't fractions, but still ... please replace 'Numerator' and 'Denominator' with something more appropriate ... such as 'premise' and 'conclusion' respectively.
OK, the rules, and some more intuitive explanation:
Universal Instantiation
'Typical' Form:
$\forall x P(x)$
$\therefore P(a)$ for any constant $a$
Explanation:
I all things have property $P$, then of course each individual thing has property $P$, whether this is $a$, $b$, ... This is why there are no restrictions here.
Universal Generalization
'Typical' Form:
$P(a)$ ... where $a$ has been introduced as some arbitrary object!
$\therefore \forall x P(x)$
Explanation:
Suppose we have a constant that we are using to denote a specific object, e.g. suppose we use the constant $c$ for 'Charlie', and suppose we have as a given that $Dog(c)$, since we know that Charlie is a dog. Now, clearly we should not be able to infer that everything is a dog just because Charlie is a dog. And that is why we mandate the constant $a$ in the rule to be a temporary name that we use to denote "some arbitrary object from our domain ... let's call it $a$" In fact, many systems require you to explicitly introduce this constant ... it would be the formal logical equivalent to the mathematician's "consider any object $a$".
I must say that in your description off the rule this requirement is not clear. ... so if you don't understand the rule as you yourself stated, I can understand that!
Here is a formal proof example:
$\forall x P(x)$ Premise
$\forall x Q(x)$ Premise
$\qquad a$ (here is where we introduce $a$ ... so we have to make sure that $a$ is not used earlier in the proof, i.e. it is a 'new' constant. Again, this is the equivalent of saying "let's consider any arbitrary object $a$. I use the indentation to create a temporary context for the use of this $a$ ... some systems use subproofs to do this)
$\qquad P(a)$ Universal Instantiation 1 (as we saw, this works for any constant, so also for $a$)
$\qquad Q(a)$ Universal Instantiation 2
$\qquad P(a) \land Q(a)$ Conjunction 4,5
$\forall x (P(x) \land Q(x))$ Universal Generalization 6 (or: 3 through 6) (so why can we do this? Because $a$ was used as an arbitrary constant!)
Existential Generalization
'Typical' Form:
$P(a)$
$\therefore \exists x P(x)$
Explanation:
Like Universal Instantiation, Existential Generalization should really be without any restrictions: If $a$ has property $P$, then there is something that has property $P$, whether $a$ is used to denote a specific or arbitrary object.
So here I am not sure why there is this restriction stated in your description of the rule...
Existential Instantiation
'Typical' Form:
$\exists x P(x)$
$\therefore P(a)$ ... for a new constant $a$
Explanation:
OK, so in this rule we do have to treat $a$ very carefully! Think about it: you know that something has property $P$ .. but do you know what it is? No. So, what the $a$ is representing here, is "some object that has property P ... which we know exists ... but we don't know what specific object it is ... so let's call it $a$". And again, like Universal Generalization, it is best to contrast the correct use of this rule with an incorrect one: Again, suppose we use constant $c$ to denote a specific individual: Charlie. Now, suppose we know that $\exists Dog(x)$ ... can we now infer $Dog(c)$? No! Because even though we know something is a dog, we don't know whether Charlie is a dog. So, like Universal Generalization, the $a$ represents an unknown object, but this time, we do know that $a$ has property $P$. And that also means that $a$ is not a completely arbitrary object .. meaning that we can't use it for a Universal Generalization.
Example:
$\exists x P(x)$ Premise
$\forall x (P(x) \rightarrow Q(x))$
$P(a)$ Existential Elimination (OK use of rule, since $a$ is a new constant)
$P(a) \rightarrow Q(a)$ Universal Instantiation 2
$Q(a)$ Modus Ponens 3,4
$\exists x Q(x)$ Existential Generalization 5
Note that we had to do line 3 before line 4, because if we would have first instantiated the universal with $a$, then we could not have instantiated the existential with that same $a$, since the $a$ is on longer a new constant!
Best Answer
The problem with replacing $\exists$ with $\neg\forall\neg$ to not have to worry about the order of quantifiers becomes apparent if you actually try doing so and omitting the quantifiers. For instance, $\exists x P(x)$ becomes $\neg \forall x \neg P(x)$ and then you omit the quantifier to get $\neg\neg P(x)$. Wait, that's equivalent to just $P(x)$, which would mean $\forall xP(x)$ under your convention. So $\exists x P(x)$ turned into just $\forall xP(x)$, which isn't right!
The problem here is that the order of negation and universal quantifiers matters. That is, $\forall x\neg P(x)$ is different from $\neg\forall x P(x)$ (so $\neg \forall x \neg P(x)$ is different from $\forall x \neg\neg P(x)$). If you omit universal quantifiers everywhere, you lose this distinction.