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.
The last paragraph makes no sense at all to me. How do you propose to read $\phi(a+3b) = 5a'+6b'$ as the definition of a function $\mathbb{Z}\to\mathbb{Z}$?
In any case, if $\phi\colon \mathbb{Z}\to \mathbb{Z}$ is a map with the property that $\phi(1) = 5$, $\phi(3) = 6$, and $\phi(2) = 2$, then $\phi$ is not an isomorphism, since $1 < 2$, but $\phi(1) = 5 > 2 = \phi(2)$, so $\phi$ does not preserve $<$.
In fact, for any integer $n$, if $\phi\colon \mathbb{Z}\to \mathbb{Z}$ is a map with the property that $\phi(1) = 5$, $\phi(3) = 6$, and $\phi(2) = n$, then $\phi$ is not an isomorphism. This is because $1 < 2 < 3$, so for $\phi$ to be an isomorphism, we'd have to have $5 = \phi(1) < \phi(2) < \phi(3) = 6$, but there is no integer strictly between $5$ and $6$.
Edit: Upon rereading, I've realized what you mean by "sums of $a + 3b$". In the structure $(\mathbb{Z};0,+)$, the substructure generated by $1$ and $3$ would be $\{a+3b\mid a,b\in \mathbb{N}\}$ (which is just $\mathbb{N}$). But note that $+$ is not in the language! The only symbol in the language is $<$, and the substructure generated by $1$ and $3$ is just $\{1,3\}$. The isomorphism $(\{1,3\},<)\to (\{5,6\},<)$ is exactly as stated: $1\mapsto 5$, $3\mapsto 6$.
Best Answer
I'm not sure about an English translation of that particular paper, but Fraïssé's book Theory of Relations is available in English translation (this includes what is today called Fraïssé theory, in Section 11.1).
I agree with HallaSurvivor's comment that Hodges' A Shorter Model Theory is the best textbook reference for Fraïssé theory.