Show that if L is regular, then third(L) is also regular. Hint: Construct an ϵ-NFA from the DFA for L
And, I suspect that that's the purpose of the stuff with the angle brackets.


This is just a way to write elements of a cross product. For example: $$\lbrace 1,2\rbrace\times\lbrace 3,4\rbrace = \lbrace \langle 1,3\rangle,\langle 1,4\rangle,\langle 2,3\rangle,\langle 2,4\rangle \rbrace$$


  I suspect that the gist of the proof is to modify the initial DFA of L with ϵ transitions, so that there are only non-ϵ transitions for every third symbol in any particular string of L, w, being analyzed.


Basically yes. The construction creates four copies of $A$. The first three copies behave just as $A$ behaved originally with a minor twist: If state $q$ is reached, instead afterwards the state $q$ in the next copy of $A$ is reached. (I will ignore the $\epsilon$ edge inscriptions for now. I come back to this later.)

Thus, initially we are in state $s_0$ of the first copy of $A$ (written as $\langle s_0,1\rangle$). Then, some letter $a$ is read. Let's say that in $A$ this would lead from $s_0$ to state $p$. Now, in the new automaton, this leads to state $p$ in the second copy of $A$, i.e. $\langle p,2\rangle$. Next, some letter $b$ is read. Let's say that this would lead from $p$ to $q$ in $A$. In the new product automaton, this leads from $\langle p,2\rangle$ to $\langle q,3\rangle$, i.e. it also moves on to the next copy of $A$.

The effect of this construction so far is that we not only "behave as $A$ would behave", but also that we know if the current letter position is exactly a multiple of three, or a multiple of three plus one, or a multiple of three plus two: All multiplies of three end up in the first copy of $A$, all multiplies of three plus one in the second copy, etc.

Now, let's go back to the $\epsilon$ edge inscriptions: So far, the automaton that was constructed just behaves as the original automaton $A$. But we want it to read just every third letter and skip the other two. We do this by replacing all edge inscriptions in the first two copies of $A$ with $\epsilon$. This will lead to a lot of non-determinism, because now all outgoing edges have the same label. The idea here is that when we have an input $a_3a_6a_9\dots$ the automaton "guesses" what the missing symbols $a_1, a_2, a_4, \dots$ could be. Each such guess is represented by taking an $\epsilon$-labelled edge. Only the third copy of $A$ is not modified, so this copy still actually reads $a_3a_6a_9\dots$


  What does the fourth copy do?


(No one actually wrote this question, but I need it for structure ;-) )

The fourth copy in the construction from the question actually does not do anything. When the automaton is in state $\langle q,4\rangle$, the only option to continue is to go to state $\langle q,1\rangle$ (via an $\epsilon$-transition). Thus, the fourth copy can just be removed and instead we go directly to $\langle q,1\rangle$.


  Can't one just add those additional transitions to a single copy of A?


First, there are no additional transitions (well, okay, the transitions from $\langle q,4\rangle$ to $\langle q,1\rangle$ can count as additional transitions). Instead, the construction keeps the original transitions but replaces some of the inscriptions with $\epsilon$.

Next, a single copy is not enough.
The automaton still has to keep track on the position in the input.
As an exercise, construct a DFA accepting $(abcd)^*$ and try out the construction on it.