First, for each $a \in \{0,1\}$ let $\bar{a} = 1+a \pmod 2$ i.e., if $a=1$ then $\bar{a}=0$, and if $a=0$ then $\bar{a}=1$.
We now consider the strategy where the 2nd player picks his bits that is the opposite of Player 1's last move is satisfied [as mentioned in the comments by Gareth Ma].
THM 1 Let $a_1,a_2,\ldots$ be a sequence of bits such that
$a_{2k+2}=\bar{a}_{2k+1}$ for each integer $k$. Then let $n$ be an odd integer, and
let $N$ be the smallest integer such that there exists an $L<N$ such that
the equation $a_{N-i}=a_{L-i}$ for each $i=0,1,\ldots, n-1$. Then $N$ is odd.
Before we establish Thm 1, we first note that Thm 1 will give us what we want. Let $a_1a_2a_3, \ldots$ be the resulting string, where, for each nonnegative integer $k$, the 1st player sets the $2k+1$-th bit $a_{2k+1}$ of the string, and the 2nd player sets the $2k+2$-th bit $a_{2k+2}$ of the string. Then if the 2nd player, for each nonnegative integer $k$, sets $a_{2k+2}=\bar{a}_{2k+1}$, then the 2nd player will end up winning, because a string of length $n$ will be repeated during the 1st player's move [because Thm 1 will give us that $N$ must be odd]. So the 2nd player has a strategy to win the game.
So we next establish Thm 1. Let us suppose that $N$ is even. We consider 2 cases:
Case 1: $L$ is odd. We make the following claim:
Lemma 2 Let us assume that the equations $a_{N-i}=a_{L-i}$ hold
for each $i=0,1,\ldots, n-1$ with $N$ is even and $L$ odd, and with $L<N$.
(i) Then with $L'=L+1$ and $N'=N-1$, the equation
$a_{N'-i}=a_{L'-i}$ holds for each $i=0,1,\ldots, n-1$.
(ii) $L \not = N-1$ so $L'<N'$.
Proof of Lema 2 To establish Lemma 2, we first note the following:
$L-n+1$ is odd, and $L-n+2j+1$ is odd for each integer $j$,
whereas $N-n+1$ is even, and $N-n+2j+1$ is even for each integer $j$.
Likewise, $L-n$ is even, and and $L-n+2j$ is even for each integer $j$,
whereas $N-n$ is odd, and $N-n+2j$ is odd for each integer $j$.
We now note the following for each $i$ satisfying $2 \le i < n$ such that $i$ is even [and $N-i$ is
even and $L-i$ is odd]:
$$a_{N-i}=a_{L-i}=\bar{a}_{L-i+1}=\bar{a}_{N-i+1}=a_{N-i+2}
=a_{L-i+2},$$
where the first $=$ comes from the equation $a_{N-i}=a_{L-i}$ for such $i$,
the second $=$ comes from the equation
$a_{2k+2}=\bar{a}_{2k+1}$ for each integer $k$, the third $=$ comes from
the equation $a_{N-i+1}=a_{L-i+1}$ for such $i$, and the
third comes from the equation $a_{2k+2}=\bar{a}_{2k+1}$ for each
integer $k$, and the fourth comes the equation $a_{N-i+2}=a_{L-i+2}$ for
such $i$. So in particular:
$$a_{N-i} = a_{L-i+2} \quad {\text{for all even $i$ satisfying $2 \le i<n$}}.$$
For $i=n$:
$$a_{N-n} = \bar{a}_{N-n+1}=\bar{a}_{L-n+1} = a_{L-n+2},$$
where the first $=$ comes from the equation
$a_{2k+2}=\bar{a}_{2k+1}$ for each integer $k$, the
second $=$ comes from the equation
$a_{N-i}=a_{L-i}$ for $i=n-1$, and the third $=$ comes from $a_{2k+2}=
\bar{a}_{2k+1}$. So in particular,
$$a_{N-n}=a_{L-n+2}.$$
This generalizes for each $i$ such that $i$ is odd [and $N-i$ is odd and $L-i$ is even] and $3 \le i < n$:
$$a_{N-i} = a_{L-i+2} \quad {\text{for all odd $i$ satisfying $3 \le i<n$}}.$$
Finally, for $i=1$:
$$a_{N-1} = \bar{a}_{N} = \bar{a}_L=a_{L+1}.$$
So from the above one can see the following:
$$a_{N-i} = a_{L+2-i} \quad {\text{for each $i=1,\ldots n$}} .$$
So then the following holds:
$$a_{N-1-i} = a_{L+1-i} \quad {\text{for each $i=0,\ldots n-1$}}.$$
From this (i) of Claim 2 follows.
To see (ii) of Claim 2, note that the equations $L=N-1$ and
$a_{2k+2}=\bar{a}_{2k+1}$ for each $k$, together imply
$a_N=\bar{a}_L$, which means that the equation
$a_{N-i}=a_{L-i}$ could not hold for $L=N-1$ and $i=0$ after all.
$\surd$
So Lemma 2 takes care of the case where $L$ is odd, so we finish next by considering the case where $L$ is even.
Case 2: $L$ is even.
Claim 3 Let us assume that the equations $a_{N-i}=a_{L-i}$ hold
for each $i=0,1,\ldots, n-1$ with $N$ and $L$ both even, and with $L<N$.
Then with $L'=L-1$ and $N'=N-1$, the equations
$a_{N'-i}=a_{L'-i}$ hold for each $i=0,1,\ldots, n-1$.
To see Claim 3, it suffices to show that $a_{N-n}=a_{L-n}$. However,
note that $N-n+1$ and $L-n+1$ are both even. So
$$\bar{a}_{N-n}=a_{N-n+1} =a_{L-n+1} = \bar{a}_{L-n},$$
and thus in particular $a_{N-n} = a_{L-n}$, and so Claim 3 follows. $\surd$
Note that Thm 1 follows immediately from Lemma 2 and Claim 3. $\surd$
Note also that the proof of Claim 3 would not carry through if $n$ were
even.
Best Answer
The $m\times n$ knight's graph is the graph whose vertices are the squares of the $m\times n$ chessboard, two squares being adjacent if a knight can move from one to the other.
Player 2 has an obvious winning strategy on the $m\times n$ chessboard if the corresponding knight's graph has a perfect matching; namely, he moves the knight to the square which is matched with the knight's current square. In particular, assuming $mn$ is even, Player 2 has a winning strategy if the knight's graph has a Hamiltonian path; in chessic terms, if a knight's tour (open or closed) exists.
Hence Player 2 has a winning strategy on the normal $8\times8$ chessboard and many others. Namely, Player 2 has a winning strategy on boards of the following sizes:
(a) $m\times n$ where $m,n\ge5$ and $mn$ is even;
(b) $2\times n$ where $n$ is divisible by $4$;
(c) $3\times n$ where $n$ is even and $n\ge4$;
(d) $4\times n$ where $n\ge2$.
For case (a) we can use the fact that a knight's tour exists. For the remaining cases, it suffices to observe that the $2\times4$, $3\times4$, and $3\times6$ knight's graphs have perfect matchings, since the boards in cases (b), (c), and (d) can be tiled with $2\times4$, $3\times4$, and $3\times6$ boards.
In a similar way, it can be shown that Player 1 wins in all other cases. That is, Player 1 has a winning strategy on boards of the following sizes:
(e) $m\times n$ where $mn$ is odd;
(f) $1\times n$ where $n\ge1$;
(g) $2\times n$ where $n$ is not divisible by $4$.
Example. Here is a "perfect matching" for the ordinary $8\times8$ chessboard:
a1c2, b1d2, c1a2, d1b2, e1g2, f1h2, g1e2, h1f2,
a3c4, b3d4, c3a4, d3b4, e3g4, f3h4, g3e4, h3f4,
a5c6, b5d6, c5a6, d5b6, e5g6, f5h6, g5e6, h5f6,
a7c8, b7d8, c7a8, d7b8, e7g8, f7h8, g7e8, h7f8.
That is, square a1 is paired with square c2, b1 is paired with d2, and so on. Note that the 64 squares are partitioned into nonoverlapping pairs, and each set of paired squares are a knight's move apart. The winning strategy for Player 2 is: WHEREVER PLAYER 1 PUTS THE KNIGHT, MOVE IT TO THE OTHER SQUARE IN THE SAME PAIR. For instance, if Player 1 starts the game by dropping the knight on e8, Player 2 replies by playing Ng7. If Player 1 now plays Ne6, Player 2 plays Ng5, and so on. Here is a sample game, with Player 1 making random moves, and Player 2 following the winning strategy described above:
P.S. Of course the strategy still works if White is allowed to move the knight to any square not already visited, and only Black is limited to making knight moves.