High Score

We can circle the numbers $1,5,7,11,2,10,4,8,0$ in that order, for a final score of $48.$ This only misses the numbers $3,6,9.$

The key fact is: if $a$ is the current number and $b$ is the next number circled, then $\gcd(a,12) | \gcd(b,12).$

So we can only circle $3$ or $9$ from a starting point of $1,5,7,$ or $11$ . Once we circle $3$ or $9,$ we can only circle $3,6,9,0$ before finishing. So the best we could do by circling $3$ or $9$ would be $1,5,7,11,3,6,9,0,$ which sums to $42.$ So the maximal-score game must avoid circling $3$ and $9.$

As for $6,$ we can circle it from anywhere except $4$ and $8$ ; but once we circle it, we can only circle $0$ and then finish. So if we circle $6,$ we can't circle $4$ and $8.$ We've already decided that we must skip $3$ and $9,$ so skipping $4$ and $8$ as well would result in a max score of $0+1+2+5+6+7+10+11=42,$ which is smaller than our already achieved score of $48.$ Hence we must skip $6$ as well. Thus $\fbox{48}$ is the maximum achievable score.