Lemma: If $A$ can win with $n=n'$, then so can $A$ win with $n=2n'$. Proof: Suppose $A$ can win with $n=n'$. Let $L$ be the set of positions over which $A$ can win in the case $n=2n'$. If $2|x-1$ then $A$ can win with the strategy for $n=n'$ by multiplying each of the original $d$'s by $2$. Otherwise, $2\nmid x-1$, so $A$ can push $x$ into $x\pm 1$ which is even. Suppose $A$ has winning strategy and $B$ places pin not $\equiv 1 \pmod p$ at the beginning. Then for all primes $p|q$ there exists a move that forces the new position to be $1 \pmod q$. As $p$ is odd, this cannot be move $1$. Thus move $2$ forces the new position to be $1 \pmod p$, but this can be done only if $\frac{p}{(k-1),p}|d$ for some $d\not \equiv 1 \pmod p$, thus $p|k-1$ for all primes $p|q$. The winning strategy for $A$: take $d = 1$ in the first move, then push with the second move until $x$ becomes divisible by $p$ and pause momentarily. This is possible since $d \equiv kd \equiv 1 \pmod p$, so we need to push $t \le p-1$ times to get to a position where $x+t \equiv 0\pmod p$. Now that $B$ had been forced to a position divisible by $p$. The point of move $1$ is to secure this added divisor $p$ to $x-1$, so that it cannot be lost again: take $d$ with added factor $p$ from now on and do this using move $1$, so that the pin is now restricted in the positions $1+pr (1 \le r \le \frac{q}{p})$. Thus we have reduced $q\to \frac{q}{p}$. Repeatedly reducing until what was $q$ becomes $1$ and we are done. Therefore the set of $n$ for which $A$ can win is exactly $\{n: \not \exists p \text{ odd prime such that } p \nmid n\}$. Thanks to user justkeeptrying for pointing out my mistake in the solution I had originally.

Call a pair $(n,k)$ good if $k$ lefts remainder $1$ when divided by the odd part of $n$. We claim that the good pairs are the only ones such that Arnaldo has a winning strategy. Step 1: If Arnaldo has a winning strategy, then $(n,k)$ is good. Proof: Assume that $(n,k)$ is not good, i.e., exists an odd prime $p$ so that $p \mid n$ and $k \not\equiv 1 \pmod p.$ We will show that Bernaldo can guarantee that the ring is never placed at a pin $1$ modulo $p$ and hence it can never be placed at the pin $1.$ First, Bernaldo places the pin at any pin which is not $1$ modulo $p.$ We will show that, whenever the ring is not placed at a $1$ modulo $p$ pin, Bernaldo can gurantees that, no matter how Arnaldo plays, in the next step the ring will not be placed at a $1$ modulo $p$ pin; which proves the claim. Indeed, assume that the ring is placed at a pin $l \not\equiv1 \pmod p.$ We have two cases: $\bullet$ If Arnaldo plays as in rule $1,$ then Bernaldo can reach his goal unless that $l+d \equiv l-d \equiv 1 \pmod p.$ But this would imply $d \equiv -d \pmod p$ and since $p$ is odd, this implies $d \equiv 0 \pmod p.$ Then $l \equiv l+d \equiv l-d \equiv 1 \pmod p$ which contradicts our hypothesis. $\bullet$ If Arnaldo plays as in rule $1,$ then again Bernaldo can reach his goal unless $l\pm d \equiv l \pm dk \equiv 1 \pmod p$ (where the two $\pm$ signs are equal); but this would imply $d\equiv dk \pmod p;$ then, since $k\neq 1 \pmod p$ and $p$ is prime, it follows that $d \equiv 0 \pmod p$ which again yelds a contradiction. $\square.$ Step 2: If $(n,k)$ is a good pair, then Arnaldo has a winning strategy. Proof: Let $n=2^ab$ where $a$ and $b$ are non-negative integers with $b$ odd. By hypothesis, $k \equiv 1 \pmod b.$ First, let $l$ be the starting pin choosen by Bernaldo. At any step, denote by $t$ the pin position. We start making $t \equiv 1 \pmod {2^a}$ in a finite number of steps. In order to do it, Arnaldo performs the following algorithm: Quote: At the first step, Arnaldo chooses $d$ so that $l$ and $d$ have different parities. So that no matter how Bernaldo plays, $t$ is now odd. Then, at the second step the base $2$ representation of $t$ is ${(d_md_{m-1}\dots d_0)}_2$ where $d_0 = 1.$ We aim to perform steps so that the first $a$ digits from right to left are $0's,$ except the right-most one. But this is very easy, whenever we have not reached our goal, select $d=2^k,$ where $k$ is the left-most non-zero digit among the first $a;$ then either Bernaldo will delete this digit(i.e. replace it by $0$) or he will move it to the left. Clearly this works. Finally, after performing the above algorithm, we will have the pin placed at a position $t\equiv 1\pmod {2^a}.$ Then Arnaldo performs rule $1$ to fix $d=2^a;$ after that, no matter how Bernaldo plays we will still have $t \equiv 1 \pmod {2^a}.$ Now, since by hypothesis $k\equiv 1 \pmod b;$ rule $2$ turns to be: Arnaldo moves the ring either clockwise or counterclowise at a number $\equiv 2^a \pmod b$ and $0\pmod 2^a.$ Hence, since $gcd(2^a,b)=1,$ if he fixes any direction and performs rule $2$ consecutively at this fixed direction, he will eventually reach a pin $\equiv 1 \pmod {b}$ and $1 \pmod {2^a};$ i.e. a pin $1 \pmod {n}.$ So Arnaldo wins!$\square.$ So expressing the possible values of $n$ as a function of $k: \{2^a b : a,b \in {\mathbb{Z}}_{\ge 0}, \text{b is odd and } b \mid k-1 \}.$