On the proof of existence of algebraic closure using Zorn's Lemma in Patrick Morandi's *Field and Galois Theory*
Any countable set can be endowed with a field structure using a bijection between the set and $\Bbb Q$ and what is known as transport of structure. Similarly, any set of size $2^{\aleph_0}$ can be given a field structure isomorphic to $\Bbb R$ or $\Bbb C$ or any other field of that cardinality.
This is similar to the way that any location on your hard drive can contain the data of this page. The set theoretic universe is just a large pagefile, so to speak. So when you talk about a field, you allocate a suitable set and endow it with the wanted structure.
In turn, this flexibility also imposes a problem. If $F$ is a field, then any set containing $F$ can be an extension of $F$ (or a subset of an extension). So when you come to apply Zorn's lemma, you have a proper class of fields to work with. But Zorn's lemma applies to sets, not to classes. So you need to somehow restrict this.
But luckily, there are only "set many" different ways to give $F$ an algebraic extension,1 since all those must have the same cardinality (plus or minus a countable set) as $F$. Therefore, we can fix a large enough set and require that all the algebraic extensions will be taken as subsets of that set, and in a coherent way.2
The reason being that all algebraic extensions can be given to a set $A$ with the same cardinality as $F\cup\Bbb N$. Each such structure is made of addition and multiplication, both of which are subsets of $A\times A$. Since there are only set many subsets to $A\times A$, there are only set many ways to give $A$ a field structure, let alone an algebraic extension of $F$.
This can be done by taking isomorphism classes and lifting the embeddings to the equivalence classes. Of course there is no a priori way of choosing representatives so that the embeddings correspond to inclusions, but we can prove using Zorn's lemma there is a maximum equivalence class which corresponds to the algebraic closure. After choosing that one, we can then choose the algebraic extensions to be ordered by inclusions as subfields of the closure.
You are overthinking this. You are correct that $S$ is just a set. An element of $\mathcal{A}$ is a subset $K$ of $S$ together with a field structure on $K$ such that this field structure makes $K$ an algebraic field extension of $F$. A subset of $S$ on its own does not automatically get a field structure, but specifying an element of $\mathcal{A}$ involves choosing one particular field structure to put on your subset.
The whole point of the Zorn's lemma argument is then to avoid the mess you describe about figuring out how to compatibly embed all possible algebraic extensions into $S$. You just pick a maximal algebraic extension of $F$ whose underlying set is subset of $S$. Then, as argued in the proof, maximality guarantees that this field is algebraically closed.