Is it just that $\forall z$ means that this statement could be true for any element,
Yes, exactly.
...and if so what's the difference between $\forall z$ and $\exists z$?
If we only asserted the existence of some particular $z$ such that if $y$ loves $z$, then that particular person $z$ would thus be $x$. But then we are not ruling out that there might be another person, different from $x$, that is also loved by everyone. And we have already asserted the existence of someone (namely, x) who is loved by all. So in that sense, $\exists z$ such that... is reduntant.
We need the universal quantifier for $z$ (to assert a statement $\forall z$) in the second clause to indicate that if there is any $z$ (which means that the claim that follows - as it relates to $z$, is true for every z) $z$ such that $L(y, z)$, then any/every such $z$ must be $x$, since there is exactly one person, namely $x$, who is loved by all $y$. I.e., for every $z,$ if every y loves z, then z must be x: i.e., that $z$ is not anyone other than $x$. This gives us that $x$ (whose existence we asserted at the start) is therefore the one unique person loved by all.
Now, just one oversight to "clean up" your expression, which you state as:
$$∃x(\forall y L(y,x))\land ∀z(∀y(L(y,z))⟹x=z)$$
But here we have two independent clauses that creates a problem, because in your second clause, you have $x$ appear outside the scope of its quantifier. I.e., it is a free, unquantified varible.
So we want the scope of $\exists x$ to persist over the entire statement, hence the square brackets below.
That is, $$\exists x \big[\forall y(L(y, x))\land \forall z(\forall y(L(y, z)) \rightarrow z = x)\big]$$
$\exists x. M(x)\implies H(x)$ reads: There exists a person such that if they are a man, they are happy.
An easier way to say this is: there is a person who is happy when they are a man.
They may still be happy when they are not a man $(F\implies T)$,
but if they are a man then they are happy $(T\implies T)$.
Likewise, when they are not happy, they are not a man $(F\implies F$),
but they are never unhappy when they are a man $\neg(T\implies F)$.
You can also do this:
$$\exists x.M(x)\implies H(x)\quad\to \quad\neg\nexists x.M(x)\implies H(x)\quad\to\quad\neg\forall x.\neg(M(x)\implies H(x))\\\equiv\neg \forall x.M(x)\land\neg H(x)$$
Not every person is unhappy when they are man.
Best Answer
First, notice that being a reindeer$-$as well as being a human$-$is a property of the individuals in the universe: you are not talking about an individual called "Reindeer". So, it is more natural to use the predicates:
The first half of the sentence ("Every human loves a reindeer") can be translated as follows:
$$ \forall x \big(H(x) \to \exists y (R(y) \land L(x,y)) \big) $$
indeed, its meaning can be pedantically reformulated as
The second half of the sentence ("Every human loves at most only on reindeer") can be translated as follows:
$$ \forall x \big(H(x) \to \forall y \forall z ((R(y) \land R(z) \land L(x,y) \land L(x,z)) \to y = z \big) $$ indeed, its meaning can be pedantically reformulated as
(To express that there is at most one individual satisfying a given property, the idea is that if two individuals satisfy such a property, then they are the same.)
Finally, the two sentences are connected by "but", whose logical meaning is equivalent to the connective "and". So, the whole translation in predicate logic of "Every human loves a reindeer, but every human loves at most only one reindeer" is:
$$ \forall x \big(H(x) \to \exists y (R(y) \land L(x,y)) \big) \land \forall x \big(H(x) \to \forall y \forall z ((R(y) \land R(z) \land L(x,y) \land L(x,z)) \to y = z \big) $$