The game has two players, player A and player B and a fixed number of rounds $n$. In each round $m$ Player A will pick an element $a_m\in\mathbb Z$ or an element $b_m\in L\times\mathbb Z$, and player B will respond by picking an element from the other model. B wins if at the end of the game the tuples $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ are locally isomorphic. If B has a winning strategy for a game of length $n$ for any $n\in\mathbb N$, then the models can be shown to be elementary equivalent.
Note that in this specific case, a local isomorphism only has to reflect the behaviour of the order: so if $a_i<a_j$, then we also should have $b_i<b_j$.
Player A can choose which model to pick elements from. Clearly if A picks from $\mathbb Z$, then B can just mirror this pick with an element from $l\times \mathbb Z$ for some $l\in L$, so we assume A picks elements from $L\times \mathbb Z$. The problem that B faces, is that A can pick elements that are infinitely far apart, while B only can pick elements finitely far apart. This is where the distance function and the number $3^{n-m}$ comes into play: B leaves enough space to accommodate for any of the possible orders in which A could pick the remaining elements.
For example, if there are four rounds, A picks $(0,0)$; B picks $0$; A picks $(1,0)$, then there are several possibilities for the remaining rounds. If $B$ picks, for example, $3$, then he gets into trouble: A can choose $(0,100)$ next, which is in between $(0,0)$ and $(1,0)$, hence $B$ needs to pick in between $0$ and $3$. But if B picks $1$, then A can pick any element between $(0,0)$ and $(0,100)$ to win, or if $B$ picks $2$, then A can pick any element between $(0,100)$ and $(1,0)$ to win. There is not enough space in between $0$ and $3$ for B to be able to answer any pick that A makes.
By choosing numbers that are at least $3^{n-m}$ apart from their closest neighbours, B makes sure to keep enough space for the remaining $n-m$ rounds of the game. I believe actually $2^{n-m}$ should to be enough: B reflects the choice of A by playing exactly in the middle of the interval in which A picked. Since there are $2^{n-m}$ numbers in the interval for each round, B can always pick a middle number and leave intervals of at least $2^{n-m-1}$ between all of the picks to accommodate for the next round.
Re: $(i)$, you write:
if the spoiler picks the vertex $c$, then the duplicator has to pick a vertex in $\mathcal{A}$ which also has two outgoing edges.
But that's not true. Remember that Duplicator wins if, at the end of the game, the atomic types of the tuples built are the same. For example, if Spoiler picks $c$, Duplicator picks $4$, Spoiler picks $a$, and Duplicator picks $3$, Duplicator will win the play. We don't check at the end of a play of the Ehrenfeucht-Fraisse game that the tuples built have matching out- (or in-)degrees.
In fact, there is very little Spoiler can do in two moves.
Best Answer
I learned about Ehrenfeucht–Fraïssé games from the wonderful, informal notes of Prof. Moschovakis, which can be found here. These notes are used in the first-year graduate logic sequence at UCLA, so they don't assume too much in the way of background. In fact, these games were covered in the first quarter of the class, so I do not believe it was necessary to be familiar with anything more than elementary first-order logic and model theory.