I'm trying to understand the following:

Thinking of a line bundle as a bunch of locally generating sections together with transition functions (which in this case are just multiplication by local sections of the structure sheaf) I wonder what it means that some power of a line bundle is trivial. I guess an example with $N>1$ would answer my question.
Thank you.
Example. If $(E,O)$ is an elliptic curve over an algebraically closed field $K$, then we know that the natural map \begin{align*} E(K) &\to \operatorname{Pic}^0(E) \\ P &\mapsto \mathcal O_E(P - O) \end{align*} is an isomorphism of groups. Thus, if $P$ is a torsion point of $E$ (they always exist: for example over $\mathbb C$ the group $E(\mathbb C)$ is a complex torus $\mathbb C/\Lambda$, for $\Lambda \subseteq \mathbb C$ a lattice, so halves of lattice points give $2$-torsion elements), then $\mathcal O_E(P - O)$ is an $n$-torsion line bundle.
Remark. If $A$ is an abelian variety over $K = \bar {\mathbb F}_p$, then every point of $A$ is torsion. This follows because any point $P \in A(K)$ is defined over a finite field $\mathbb F_q$, and the group $A(\mathbb F_q)$ is finite.
In particular, applying this to $\operatorname{Pic}^0(X)$ for any smooth projective variety $X$ shows that every line bundle rationally equivalent to $\mathcal O_X$ on a smooth projective variety over (the algebraic closure of) a finite field is torsion. Presumably the citation you gave uses a variant of this statement, for varieties over $\mathbb Z/p^n\mathbb Z$.
Remark 1 (thanks to Alex Youcis). I of course have to assume something like $\mathscr L$ is rationally equivalent to $\mathcal O_X$, because otherwise the result I claim is clearly false: think for example about $\mathcal O_{\mathbb P^n}(1)$.
Remark 2. Because Alex Youcis asked: the precise statement about representability of $\operatorname{Pic}^0_X$ for flat projective $X \to \operatorname{Spec} \mathbb Z/p^n\mathbb Z$ with geometrically integral fibres is proven in FGA Explained, Theorem 9.4.8. But this is probably much beyond the type of material the OP is comfortable with.
Short story: it works also over $\mathbb Z/p^n\mathbb Z$, but my proof is fairly elaborate. There might be a much easier argument that I'm unaware of, but this is the standard answer that most algebraic geometers will pull out.