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 main thing to note here is that this is analogous to the game where one has as many of each card as desired, rather than just four. In particular, it is easy to see that, in this modified game, the winning positions are exactly the positions where the sum is of the form $31-7n$ for some $n$. This is presumably what is meant by the "target series". Therefore, if you play $5$ and your opponent plays $3$, then your next move should be to play $2$, not $4$, since $2$ brings the sum of all the flipped cards to $10=31-7\cdot 3$. That is, the strategy is as follows:
I think the misunderstanding is in what it means to enter the "target series". In particular, you seem to have understood this as meaning that a player should always make sure that the sum of their move and their opponent's last move is equal to $7$. While it is true that this will happen once you are in the target series, in order to move from not being in the series to being in the series, some other sum is desired.