Let's start with the given wff as that seems to be causing difficulty, and work towards its translation into English. Then we've reverse the process!
$$\exists y(\forall xF(x,y)\land(\forall z((\forall wF(w,z))\rightarrow y=z))$$
It can help a great deal to go from logic to English (or vice versa) in stages, via "Loglish" -- that unholy mixture of English and symbolism which we cheerfully use in the classroom! So ....
There is someone $y$ such that $(\forall xF(x,y)\land(\forall z((\forall wF(w,z))\rightarrow y=z))$
can be read
There is someone $y$ such that (everyone $x$ is such that $x$ can fool $y$) and (everyone $z$ is such that ($(\forall wF(w,z))\rightarrow y=z))$
i.e.
There is someone $y$ such that everyone can fool $y$ and everyone $z$ is such that (if everyone $w$ can fool $z$, then $z$ is the same person as $y$).
i.e.
There is someone $y$ such that everyone can fool $y$ and anyone whom everyone can fool is none other than $y$ again.
i.e.
There is someone whom everyone can fool, and no one other then he can be fooled by everyone.
i.e.
There is exactly one person whom everyone can fool.
Read this from top to bottom to translate the formal wff into English.
And now read the same sequence from bottom to top to translate in the other direction!!
Taking things in stages like this helps a great deal when first learning to translate in either direction. There are lots more worked examples of this kind involving nested quantifiers in my Introduction to Formal Logic (Ch. 24), with more exercises and answers online. For practice quickly makes perfect: but it does take a bit of practice to make this all seem as easy as it really is. I recall Paul Teller's A Modern Formal Logic Primer is also quite good on translation (his book, now out of print, is freely available from his website).
Some comments:
(i) In propositional logic, a propositional variable stands for a sentence, i.e. something that can be true or false.
This is the reason why, starting from a first-order formula, we have to "remove" free variables; a well-formed f-o formula like "$x$ is green" (i.e. $green(x)$) is not a sentence, because we cannot say, without assigning a reference to $x$ if it is true or false.
Thus, we have to do one of the two thing mentioned by you: either replace $x$ with a constant ( a "name") or bind the variable with a quantifier.
(ii) $\forall x green(x)$ is a prefectly understandable sentence: "all objects are green".
You are right in saying that in a finite domain, $\forall x green(x)$ is equivalent to : $green(c_1) \land \ldots \land green(c_n)$.
But with first-order logic we are interested to "handle" also infinite domain; and in mathematical logic, we are interested primarily to handle with infinite domain.
And we do not (usually) admit as well-formed formulae infinite long expressions (but see Infinitary Logic).
If we restrict ourselves to finite domain, an universally quantified formula is a (finite) conjunction, while an existetially quantified formula is a (finite) disjunction.
Thus, if we restrict ourselves to finite domain, we can simply dispense with quantifiers ...
(iii) Having said that, what is the purpose of "downsizing" a first-order formula to a propositional one?
This technique can be useful in order to show that a f-o formula is valid: if the "propositional translation" of a f-o formula is a tautology, we are sure that the original f-o formula is valid.
Consider for example the (obviously) valid formula:
$\forall x (x=0) \lor \lnot \forall x (x=0)$;
we can translate it as $p \lor \lnot p$, that is a tautology.
Thus the original f-o formula is valid.
But we have valid formulae that are not "instances" of tautologies; consider :
$\forall x (x \ge 0) \rightarrow (0 \ge 0)$.
Its "translation" is $p \rightarrow q$, that is not a tautology.
This fact must be obvious; first-order language as a greater "expressive power" than propositional language; with f-o language we can perform a "deeper" decomposition of the logical structure of sentences (compared to the "decomposition" that we can perform only in term of the conncetives).
Thus, we are able to "discover" more logical truth than with "propositional decomposition".
Best Answer
There are two issues here. One is that the logic statement you offer does not match the English statement. The logic statement you offer (when you add $\forall y$ as Mauro Allegranza suggested in the comments) would be best translated to English as: The sum of any two positive numbers is (necessarily) positive.
For the English statement "The difference of two positive numbers is not necessarily positive", you would want something like: $$\neg \forall x \forall y (x>0 \land y>0 \rightarrow x-y>0)$$ or, equivalently, $$\exists x,y \text{ such that } (x>0 \land y>0 \land x-y <0)$$
The second issue is the question you raise about how "necessarily" figures in these statements. "Necessity" is a phrase used in many ways, often unclearly. For your purposes, in your original English sentences, "necessarily" almost certainly means simply that some positive numbers have negative difference.
In other words, the only reason "necessarily" is there in your English statement is because if it weren't there, many readers might erroneously understand the statement to be claiming that all positive numbers have negative differences. But the statement is not claiming that; it is instead claiming that positive numbers sometimes have negative differences. So "necessarily" is added to make this clear.
The reason why "necessarily" does not occur in your second example is that there is no such danger of ambiguity, so there is no reason to include "necessarily" to prevent misunderstanding.
It is worth adding that some "modal" logical systems -- almost certainly not related to what you are working with -- do incorporate operators for necessity. In such a logic, there is a distinction between truth and necessary truth. (E.g., it is true that my tea is hot, but it is not a necessary truth; my tea can easily become cold. By contrast, it is a necessary truth that 1<2.) In such modal logical languages, necessity is typically denoted by "$\square$". So, we have:
\begin{align*} A \text{ implies }B& & \phantom{\square (}A\rightarrow B\phantom{)}\\ A \text{ necessarily implies }B& &\square (A\rightarrow B) \end{align*}