Xjyuk

The traditional brute force collision attack is generate $2^N/2$ (unique) random strings, hash them and this results in ~50% chance for collision.
The attack talked in the question's title is generate hashes by the sequence $H_0 = F(S), H_n = F(S||H_n-1)$, where $S$ is a string to make collision with, $F$ is the hash function, and the sequence stops at $2^N/3$.

The above attack needs less calls to the hash function because it relies on the fact that each sequential hashing reduces the number of possible hashes. When hashing $2^N$ random strings, there will be $(1-(1-frac12^N)^2^N)2^N$ possible hashes because of all the collisions.

The sequence that tells the fraction of the possible hashes for each hash in the sequence is $P_0 = 1, P_n = 1 - (1-frac12^N)^2^NP_n-1$, which roughly approximates to $P_n = P_n-1 - fracP_n-1^22$ as $lim_P_n-1to 0$, which can be approximated into a function $P(n) = frac2n + 2$. (these reductions in precision make it easier for future calculations)

The chance for a collision to NOT happen for a hash in the sequence against all of it's following hashes is $C_n = (1 - frac1P_n2^N)^M - n - 1$ where $M$ is the sequence's length, because the fraction is so close to $1$, it can be approximated to $C_n = 1 - fracM - nP_n2^N = 1 - fracMn - n^22^N + 1$. Now to approximate the multiplication of all $C$s ($C_0C_1...C_M)$ to get the chance for collision for all hashes combined, we can discard the multiplication (because again, $C_n$ is very close to $1$) and add all the subtracted parts and subtract them from $1$, which results in $1 - fracM^3/2-M^3/42^N+1 = 1 - fracM^3/42^N+1$, and to reach ~50% collision rate $M = 2^(N+3)/3$, which is roughly just $2^N/3$.

Is there a reason why this won't work?

It is correct that the set of possible $H_n$ over all the possible $S$ reduces as $n$ grows. However the attack evaluates the $H_i$ for a fixed random $S$, not for multiple $S$; thus that reduction is immaterial to the success of the attack.

In other words: it is evaluated the number of possible $H_i$ for random $S$ as a function of $i$, and from that drawn conclusions for collision probability among the $H_i$ for sequential $i$ and a particular $S$. The conclusions are thus not justified, and, it happens, quite incorrect.

When we model $F$ as a random function, then until there is a collision the $H_i$ are random, and hence the probability of collision among the $H_i$ with $0le i<n$ is per the birthday bound.

Let $M$ be the number of queries to a uniform random function $F$ at distinct points $X_1, X_2, dots, X_M$. The probability of a repeated value $F(X_i) = F(X_j)$ for $i ne j$ is at most $M^2!big/2^N$ by the birthday paradox. For $M = 2^N/3$, this is $2^2N/3!big/2^N = 1big/2^N - 2N/3 = 1big/2^N/3$. If your calculation doesn't fit this bound, there's an error in your calculation.

‘But what if we come upon a short cycle in $H mapsto F(S mathbin| H)$?’, you ask. The expected cycle length of a uniform random function is $frac12sqrt2pi,2^N approx 2^N/2$ (Harris 1960 Eqs. 3.2, 3.4, 3.11; paywall-free), so no, this method doesn't improve the expected cost.

Using a clever tortoise-chasing-hare algorithm to find a cycle, rather than making a big table and checking for duplicates, may reduce the memory or area*time cost of an attack, as in the van Oorschot–Wiener parallel collision search machine, but it doesn't escape the birthday bound!