The sketch is somewhat formally problematic; it is intuitively appealing, but it would take a fair amount of formal machinery to actually get it work. That’s part of where “ordinals” come in.
To answer your questions: yes, that is the negation of Zorn’s Lemma. You can write Zorn’s Lemma as:
If
(for all $T\subseteq P$
(If $T$ is totally ordered,
then there exists $y_T\in P$ such that $x\leq y_T$ for all $x\in T$))
then
(there exists $m\in P$ such that
(for all $y\in P$
(if $m\leq y$
then $m=y$))).
The negation of “if $P$ then $Q$” is “$P$ and not($Q$)”. So the negation of Zorn’s Lemma would be
For all $T\subseteq P$
(if $T$ is totally ordered,
then there exists $y_T\in P$ such that $x\leq y_T$ for all $x\in T$)
and
(for all $m\in P$
(there exists $y\in P$ such that
$m\leq y$ and $m\neq y$)).
The last, “$m\leq y$ and $m\neq y$” is the same as $m<y$. So the latter clause reads: for every element of $P$ there is an element of $P$ that is strictly greater than it.
The idea of the proof is: since the empty set is a totally ordered set and has an upper bound, $P$ is not empty. Say $a_1$. But $a_1$ is not maximal, so there exists $a_2>a_1$. But $a_2$ is not maximal, so there exists $a_3>a_2$. “Continuing this way” (Choice, since it involves infinitely many choices) we get $a_1<a_2<a_3<\cdots$; let $T=\{a_1,a_2,\ldots\}$. This is totally ordered, so it has an upper bound $a_{\omega}$. This is not maximal, so there is $a_{\omega+1}>a_{\omega}$. “Continuing this way” (Choice) you keep going and going and going and going. Using some technical set theoretic machinery, you can then guarantee that you “keep going” past countability and really past any set size. So this would allow you find a collection of elements of $P$ that is strictly larger (in the sense of cardinality) than $P$ itself. This is of course impossible, leading to the contradiction.
The “ordinals” are types of well-ordered sets, and they are the generalization of natural numbers. They allow you to define “inductions” that go past the natural numbers and through to other infinite well-ordered sets as indices.
If you want a different version of the proof, I recommend George Bergman’s handout, The Axiom of Choice, Zorn’s Lemma, and all that, available from his website
We first show that $U$ is a chain. It’s not hard to see that $U$ is a chain. Suppose that $x,y\in U$; there are conforming sets $A_x$ and $A_y$ such that $x\in A_x$ and $y\in A_y$. Lemma $1$ implies that one of these sets is a subset of the other, so without loss of generality suppose that $A_x\subseteq A_y$. Then $x,y\in A_y$, and $A_y$ is a chain, so either $x\le y$, or $y\le x$.
Now we show that $U$ is well-ordered by $\le$. Suppose that $\varnothing\ne S\subseteq U$; we want to show that $S$ has a $\le$-least element. Fix $s\in S$; there is some conforming set $A$ such that $s\in A$. $A$ is well-ordered by $\le$, so let $a=\min\{x\in A\cap S:x\le s\}$; I claim that $a=\min S$. Let $x\in S$; if $x\in A$, then certainly $a\le x$, so suppose that $x\in S\setminus A$. There is some conforming $A_x$ such that $x\in A_x$, and clearly $A_x\nsubseteq A$, so $A$ is an initial segment of $A_x$. Since $x\in A_x\setminus A$, it’s clear that $a<x$ in this case as well and hence that $a=\min S$.
It remains to be shown that if $x\in U$, then $x=f(P(U,x))$. Let $x\in U$; there is a conforming set $A_x$ such that $x\in A_x$. By definition $x=f(P(A_x,x))$, so we’re done if we can show that $f(P(U,x))=f(P(A_x,x))$. It suffices to show that $P(U,x)=P(A_x,x)$. Clearly $P(A_x,x)\subseteq P(U,x)$, so we need only show that $P(U,x)\subseteq P(A_x,x)$, i.e., that if $y\in U$ and $y<x$, then $y\in A_x$. Suppose, then, that $y\in U$ and $y<x$; there is a conforming set $A_y$ such that $y\in A_y$. If $A_y\subseteq A_x$, then certainly $y\in A_x$, and we’re done. Otherwise, $A_x$ is an initial segment of $A_y$, and again $y\in A_x$, since $y<x$.
Best Answer
No, the latter condition is strictly stronger. For example, consider the set $[0, 1]$ ordered by the usual ordering on the reals. Note that $[0, 1]$ has a maximal element $1$ (in fact, the maximum), so indeed every totally ordered subset has $1$ as an upper bound. However, not every totally ordered subset has a maximum. In particular, $(0, 1)$ has no maximum.