Denote by $(a, b, \sigma)$ the game state where you have $\$a$, your opponent has $\$b$, and your score is $\sigma$. Clearly the game is scale-independent, so the ratio of your money to your opponent's is all that matters to determine the winner. Also, clearly it can only help you to have more money at any point; so there must be exactly one transition (as the ratio is increased) from losing to winning for each score. Let $R_\sigma$ be the critical ratio when your score is $\sigma$; that is, if the ratio is larger, you win, and if the ratio is smaller, you lose. By symmetry, $R_0=1$. (The winner at the critical ratio for each score could still go either way; certainly it depends on the tie-breaker rules, while the rest of this analysis does not.)
First consider $\sigma=+2$. You win games $(r,1,+2)$ with $r > 1$; for $r \le 1$, your opponent can (must) match your bet, which should therefore be as large as possible, leading to the game $(r,1-r,+1)$, which goes from losing to winning at $r/(1-r)=R_1$. So $R_2/(1-R_2)=R_1$, or $R_2=R_1/(1+R_1)$.
Now consider $\sigma=+1$; you again clearly win for $r > 1$, but for $r \le 1$, your opponent can choose whether to match your bet or not. Let $s(r) \le r$ be your bet. Then your opponent can choose whether the next state is $(r-s(r), 1, +2)$ or $(r, 1-s(r), 0)$. The latter is a loser for you if $r < 1-s(r)$, so you must have $s(r) \ge 1-r$. The former is a loser for you if $r-s(r)< R_2$, so you must also have $s(r) \le r-R_2$. These conditions become incompatible (that is, your opponent can always win) if $1-r>r-R_2$, or $r<(1+R_2)/2$. We conclude that $R_1=(1+R_2)/2$. Combining this with the previous relation between $R_1$ and $R_2$, we find that $R_2=-1+\sqrt{2}$, and that $R_1=\frac{1}{2}\sqrt{2}$.
Finally, let's determine your optimal bet from $(1,1,0)$. If you bet $s$, your opponent can choose whether the outcome is $(1-s,1,+1)$ or $(1,1-s,-1)$. One of these is a clear loser unless $1-s=R_1$, or $$s=1-R_1=1-\frac{1}{2}\sqrt{2}.$$
This is your optimal initial play.
Note that because of the unfair tiebreaker rules, your opponent will win this game... if you bet exactly the critical amount each round, then your opponent will exactly match your bets and win in three rounds. However, there is a silver lining: you can draw out your defeat, making your opponent spend an arbitrarily large number of rounds reaching her inevitable victory. This can be done by betting your opponent's wealth plus a sufficiently small $\varepsilon$ when the score reaches $-2$.