For every element in $S$ there is a choice to be made: it belongs
to $A-B$, to $B-A$ or to $A\cap B$. That gives $3^{n}$ possibilities
where $n$ denotes the cardinality of $S$.
However, this result corresponds
with the cardinality of $\left\{ \left\langle A,B\right\rangle \mid A\cup B=S\right\} $
and what we are really after is the cardinality of $\left\{ \left\{ A,B\right\} \mid A\cup B=S\right\} $.
If $A\neq B$ then $\left\langle A,B\right\rangle \neq\left\langle B,A\right\rangle \wedge\left\{ A,B\right\} =\left\{ B,A\right\} $
and a double counting must be repaired. In special case $A=B$ the
condition $A\cup B=S$ leads to $A=B=S$ and only in that case there
is no double counting.
This leads to $\frac{1}{2}\left(3^{n}-1\right)+1$
for the cardinality of $\left\{ \left\{ A,B\right\} \mid A\cup B=S\right\} $.
I think this is the issue:
An $(8, 5, 4)$ covering design is a set $\mathcal{B}$ of subsets of $\{1, 2, \ldots, 8\}$ (hereafter I'll use $[8]$ to refer to $\{1,2, \ldots, 8\}$) such that any $4$-element subset of $[8]$ is a subset of some $B \in \mathcal{B}$.
What you have is not an $(8, 5, 4)$ design: for example, $\{1, 2, 4, 7\}$ isn't a subset of any of your five $5$-element subsets of $[8]$.
But, your sets are enough to guarantee (and I had Sage help me make sure!) that for all $5$-element subsets of $[8]$, at least one of its $4$-element subsets is a subset of one of your sets.
This is not my area of expertise, so I'm not aware of any good references for the sort of things you're looking to construct. But, they are not $(v, k, t)$ covering designs.
From the answer to your other question:
A Covering Design $C(v,k,t,m,l,b)$ is a pair $(V,B)$, where $V$is a set of $v$ elements (called points) and $B$ is a collection of $b$ $k$-subsets of $V$ (called blocks), such that every $m$-subset of $V$ intersects at least $l$ members of $B$ in at least $t$ points. It is required that $v \geq k \geq t$ and $m \geq t.$
You want $l=1,k=5,$ $V=\{1,2,\ldots,14\} \quad(v=14)$ etc. Google the "La Jolla Covering Repository" for extensive tables.
We can reconcile this with the definition given by La Jolla:
A $(v,k,t)-$covering design is a collection of $k$-element subsets, called blocks, of $\{1,2,\ldots ,v\}$, such that any $t$-element subset is contained in at least one block.
If we are talking about a $(v, k, t)$ covering design, it fits into the above definition, but it's suppressed some parameters: Namely, requiring every $t$-element subset to be contained in at least one block, we're stipulating that $m = t$ (in your case, $4$).
Specifically, your sets are a
$$C(v = 8,\, k = 5,\, t = 4,\, m = 5,\, l = 1,\, b = 5)$$
covering design. But what's listed in La Jolla is a
$$C(v = 8,\, k = 5,\, t=4,\, m=4,\, l=1,\, b=20)$$
covering design. I suspect that you need $b \ge 20$ subsets to cover $m = t = 4$-element subsets of $[8]$ (hence the $20$ sets in the repository) while, evidently, covering $m = 4$-element subsets of all $t = 5$-element subets of $[8]$ requires fewer sets; at most $b = 5$, as your collection shows.
Best Answer
What Frentos wrote in the comments is incorrect. The reverse lexicographical order on subsets of $[12]$ with five elements is found by pretending that $12<11<10<\ldots<3<2<1$ and using the resulting lexicographical order. Thus, the first two are indeed $\{12,11,10,9,8\}$ and $\{12,11,10,9,7\}$, but the next one after that is $\{12,11,10,8,7\}$, not $\{12,11,10,9,6\}$.
This means that your aside is not correct. It would be almost correct if you were talking about the reflected lexicographic order, which is simply the reversal of the usual lexicographic order: it would be off by one, since the $233$-rd element from the end of the lexicographical order is the $(79\color{red}3-233)$-rd, or $560$-th element, not the $559$-th element. Here’s a HINT for the correct approach: if $a<b<c<d<e$, compare the position of $\{a,b,c,d,e\}$ in the lexicographic order with the position of $\{13-a,13-b,13-c,13-d,13-e\}$ in the reverse lexicographic order.