Answer. Ariane wins for $n = 2, 4$ and $8$, for all other $n \ge 2$ Bérénice wins.

We observe: If Ariane can win for a certain $n$, she will also win for all divisors of $n$, and conversely, if Bérénice can win for a certain $n$, she will also win for all multiples of $n$ because a residue $0$ modulo $n$ is automatically a residue $0$ for all divisors of $n$.
It remains to show that Ariane wins for $n = 8$ and Bérénice wins for $n = 16$ and $n$ odd.
All congruences in this solution are modulo $n$.

* For $n = 8$, Ariane has to choose $2$ in the first step. If Bérénice takes $4$, Ariane can choose $8 \equiv 0$ and has won. If Bérénice takes $3$, Ariane can choose $6$. Now, Bérénice has to decide between $7$ and $2 \cdot 6 = 12 \equiv 4$. But for both, Ariane can immediately choose $8 \equiv 0$.

* For $n = 16$, Bérénice chooses $2x$ for all numbers except $4$ and $8$. This clearly never gives the residue classes $0$, $15$ or $8$, so that Ariane also cannot choose $0$.

* For $n = 3$, Ariane has to choose $2$ in the first step and then Bérénice chooses $1$ again, which means that Bérénice wins.

* For odd $n > 3$, it is not possible to reach $0$ with $2x$ from another residue class. So the only possible issue for Bérénice would be the situation that both her options are among $n$ and $n-1$ such that she or Ariane choose $0$. But this means that $x+1$ takes the residues $0$ or $-1$, so $2x$ takes the residues $-2$ or $-4$ which are both different from $0$ and $-1$, so this cannot happen and Bérénice can permanently avoid $0$ being chosen.