Disclaimer: A lot of this post was copied from my answer to a similar sort of game.
Summary
It turns out that this game is very closely related to the Octal Game with code "0.007". Using the Sprague-Grundy Theorem combined with a winning strategy for Nim, it is relatively straightforward to analyze small games like this (e.g. your game for small $n$), and there are theorems which give efficient algorithms to learn about this in the medium range. Some Octal games can be efficiently solved even in large cases, but this particular game is not one of them.
It turns out that $B$ has a winning strategy for $N={0, 1, 2, 8, 14, 24, 32, 34, 46, 56, 66, 78, 88, 100, 112, 120, 132, 134, 164, 172, 186, 196, 204, 284, 292, 304, 358, 1048, 2504, 2754, 2914, 3054, 3078, 7252, 7358, 7868, 16170}$ and $A$ has a winning strategy in all other cases up to $2^{28}-1$. But I believe the general result is still open.
Connecting this game to existing theory
First note that you shouldn't be able to draw a curve through points as then drawing a curve through all of the points would be a winning move all of the time. To connect this game to existing general theory, we can reconceptualize it in the way that Lærne did: After you draw a curve, you remove three points from a component, and split the remainder into two components (where one or both of those may be empty). That is traditionally encoded in a "number" with Octal digits. In this case, the code is $0.007$, where the $0$s indicate that you can't remove only one or two points, and the three bits of the $7$ indicate that you can leave zero or one or two components when you remove three points from a component.
Who wins?
Since the original question was "who has a winning strategy?", rather than "what is the winning strategy, I won't re-present an introduction to the relevant theorems in detail.
If you don't already know about the strategy for Nim or the Sprague-Grundy Theorem and how they can be applied to similar games, you can read one or more of the following:
Very briefly, positions in games like this act in combinations like single heaps of Nim, where the size of the corresponding Nim heap is the least number that's not a size of a Nim heap corresponding to a position you can move to. The strategy for Nim tells you that combining games with known equivalent Nim boils down to bitwise XOR (aka "adding in base $2$ without carrying").
Because of the strategy for Nim (how Grundy values combine), and a position in this game is a combination of separate components, it suffices to know the Grundy/Nim values for a single "heap" (in our context, a single component of points). There are tricks to make this calculation more efficient than the naive method, some of which are covered in the graduate textbook "Combinatorial Game Theory" by Aaron N. Siegel. For some similar games, the Nim values are known to be eventually periodic (by a theorem that says they will be if they look periodic long enough), but according to Flammenkamp, for $0.007$, $2^{28}$ values have been calculated with no periodic behavior, although the values $N$ for which $B$ wins appear to stop at $16170$.
Aaron Siegel's program Combinatorial Game Suite allows efficient calculation of middling values (but probably wouldn't get up to $2^{28}$. For example, the lines hr:=game.heap.HeapRules.MakeRules("0.007") and hr.NimValues(100000) takes about a minute and a half on my machine to produce the list of Nim values for $N=0$ to $N=99999$.
For example, the Nim values from $N=0$ to $N=499$ are as follows (remember that $N=0$ corresponds to a win for $B$) $0,0,0,1,1,1,2,2,0,3,3,1,1,1,0,4,3,3,3,2,2,2,4,4,0,5,5,2,2,2,3,3,0,5,0,1,1,1,3,3,3,5,6,4,4,1,0,5,5,6,6,2,7,7,7,8,0,1,9,2,7,2,3,3,3,9,0,5,4,4,8,6,6,2,7,1,1,1,0,5,5,9,3,1,8,2,8,5,0,1,1,12,2,2,7,3,3,9,4,4,0,11,3,3,3,9,2,2,8,1,3,5,0,9,12,2,6,13,13,5,0,1,1,4,11,7,7,10,3,4,1,4,0,5,0,3,3,6,7,2,14,13,10,4,12,9,2,2,3,3,6,9,9,1,16,4,8,3,3,2,15,1,1,4,0,5,5,16,6,6,6,8,0,16,5,4,4,17,2,2,7,14,6,10,12,1,0,16,13,3,6,2,7,7,8,1,0,5,17,2,12,15,3,11,0,19,18,12,4,16,17,2,2,21,6,9,4,19,5,5,17,10,3,6,19,2,7,8,4,1,9,12,7,2,13,6,3,19,5,9,4,8,8,17,17,2,15,18,1,1,8,5,21,16,21,3,19,19,13,5,18,1,4,17,7,2,7,6,3,19,12,5,5,16,16,6,17,19,7,7,18,1,4,17,0,9,16,3,3,14,13,22,0,1,15,24,17,2,6,18,3,4,19,19,0,8,21,16,3,15,7,26,18,13,1,1,17,9,2,21,2,6,22,19,9,5,16,4,16,20,3,7,18,23,22,8,20,5,16,21,15,6,10,19,18,18,18,4,4,17,17,7,2,3,23,19,9,5,0,16,16,3,17,30,2,18,18,8,4,17,17,9,27,6,10,19,19,14,9,9,4,20,17,14,11,7,18,6,19,19,5,13,16,16,10,6,19,19,23,18,4,4,17,12,12,14,10,6,3,19,5,9,5,21,16,20,6,7,7,18,30,13,13,17,12,21,15,10,3,19,22,18,8,4,32,17,17,11,14,6,26,24,12,5,9,16,16,6,7,7,7,18,18,8,4,17,20,7,16,10,10,22,19,22,9,23,4,13,17,20,7,11,23,23,4,5,9,5,16,16,10,17,10,22,18,23,8,4,17,17,20,16,32,13,19,19,33,5,5,24$
Best Answer
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.