If the moves consist of taking either $a$ or $b$ sticks, with $a\lt b$ and, as clarified in a comment, $N\lt a$ is a loss, then the first player wins iff $((N-1)\bmod(a+b))\bmod(2a)\lt a$, and a winning strategy is to take $a$ sticks if $(N-1)\bmod(a+b)\gt b$, take $b$ sticks if $(N-1)\bmod(a+b)\lt a$, and take either otherwise.
Claim 1: If there are two contiguous groups of coins left of equal size, then the player to move loses.
Proof: If the player makes a certain move in one group, the other player can mimick the same move in the other group. At some point the first player will eliminate one of the groups, at which point the second player can also remove the other group, winning the game.
Claim 2: If there is one (non-cyclic) chain left of any length, then the player to move always wins.
Proof: If there is an even number of coins in the chain, you remove the middle $2$ coins, and Claim 1 then proves the result. Similarly, if there is an odd number of coins, you remove the middle coin, and use Claim 1 to finish the proof.
Claim 3: If there is one cyclic chain left of any length $\geq 3$, then the player to move always loses.
Proof: Removing any number of coins from anywhere in the cycle always leaves a non-cyclic chain of a certain length, allowing the next player to win by Claim 2.
So I think the final answer is that the player to move always loses, unless there are only $1$ or $2$ coins in a "circle".
Best Answer
A hint:
Write the numbers from $1$ to $25$ or so in a row and below each number $k\geq1$ recursively write a $W$ if you are sure that $k$ matches left is a winning position, and an $L$ if you are sure that $k$ matches left is a losing position: $$\matrix{1&2&3&4&5&\ldots&25 \cr W&L&\ldots\cr}$$ A position is winning if you can put the opponent into a loosing position by taking an appropriate number of matches, and is losing if all allowed moves lead to a winning position for the opponent.
When you don't see the pattern by then replace $25$ by $60$ or so.