Correctness of AKS algorithm for shortest vector problem
What you seem to be missing is that $\tau$ is not applied to all "green" vectors. Instead, think of every point $x_i$ as having a coin attached to it. Before you use $x_i$ in the algorithm, you toss the coin to decide whether to keep $x_i$ as $x_i$ or to replace it by $\tau(x_i)$. This is done independently for each $x_i$, so, with very high probability, the fraction of green vectors that $\tau$ is applied to will be very close to $\frac12$. 

I have always found this analysis mysterious, but let me try to give some intuition. I think the analysis tries to exploit the fact that the sieving step only uses $y_i$, and $y_i$ does not distinguish between $x_i$ and $x_i\pm v$ for any lattice vector $v$. Since the sieving step does not need this information, you can think of the analysis as "hiding" from the algorithm which point in the pair $\{x_i, \tau(x_i)\}$ was actually sampled, until the algorithm absolutely must know this information. This is sometimes called the principle of deferred decisions. With this trick, you can argue that there are many possible random choices that would've led to the same set of points surviving to step 3 of the algorithm, and then you argue that at step 3 many of these random choices lead to a correct output. On the other hand, if you applied $\tau$ to every point, then you wouldn't "hide" anything from the algorithm, i.e. you wouldn't be taking advantage of the fact that there is randomness that's not used by the sieving step 2, and is "free" to be used at step 3.