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 ($(\forall wF(w,z))\rightarrow y=z))$
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 $y$).
There is someone $y$ such that everyone can fool $y$ and anyone whom everyone can fool is none other than $y$ again.
There is someone whom everyone can fool, and no one other then he can be fooled by everyone.
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).