You were on track at first, but the game may go on for a very long time. You should not stop at "$2n$".
$A$ could win on the first flip of the coin, or the third, or the fifth, ...
So, the probability of $A$ winning would be the sum of the probabilities of the events
$$
A_i = A\text{ wins on the } i^{\rm th}\text{ flip; where }i\text { is odd}.
$$
If the coin is fair, $$P(A_i)
=\Bigl({1\over2}\Bigr)^{i-1}\cdot{1\over2} =\Bigl({1\over2}\Bigr)^i.
$$
Summing the probabilities above, the probability that $A$ wins is
$$
\sum_{i \text{ odd}} P(A_i)= \sum_{i \text { odd}} \Bigl({1\over 2}\Bigr)^i.
$$
The series above can be written:
$$
\sum_{i \text { odd}} \Bigl({1\over 2}\Bigr)^i=
\sum_{i =0}^{\infty} \Bigl({1\over 2}\Bigr)^{2i+1}=
{1\over 2}\sum_{i =0}^{\infty} \Bigl({1\over 4}\Bigr)^{ i }={1\over2}\cdot{4\over3}=2/3.
$$
You could also solve the problem this way:
Condition on what happens on the first two flips:
$A$ wins if the first flip is a head and the probability that the first flip is a head is 1/2.
$A$ loses if the first flip is a tail and the second ($B$'s turn) is a head. The probability that the first flip is a head and the second a tail is 1/4.
If the first two flips are tails, then given this, the probability that $A$ wins eventually afterwards is the same as the initial probability that $A$ wins. The probability that the first two flips are tails is 1/4.
So
$$
P(A) =1\cdot{1\over2}+0\cdot {1\over4}+P(A)\cdot{1\over4}.
$$
Solving the above for $P(A)$ gives $P(A)=2/3$.
As for your second method, I think you are forgetting that $A$ does not get a second toss if $B$ flips heads on his first toss...