This is so that there is no "conical singularity" at the horizon (not the origin). Let me elaborate.
In general, how do you know if a spacetime has a curvature singularity? Well if any metric components diverge or vanish at any point, then you know you might be in trouble. However you still need to determine if you have a curvature singularity (i.e. a true singularity) or a coordinate singularity (i.e. it was just a bad coordinate system you chose). The failsafe way to determine if it's merely a coordinate singularity is to find a coordinate system where the metric is regular at the point of interest.
For the Schwarzschild metric, we know the singularity at $r=2M$ is a coordinate singularity because there exists a coordinate transformation where the metric is smooth at that point (Kruskal coordinates or Eddington-Finkelstein coordinates).
Similarly, if you were faced with a metric
$$ ds^2 = dr^2 + r^2 d\theta^2$$
you might be concerned that the space is singular at $r=0$. However you know that this is merely a coordinate singularity because if you do the coordinate transformation $x = r \, \cos\theta$ and $y=r\, sin\theta$ the metric is now
$$ds^2 = dx^2 +dy^2$$
and is regular everywhere.
However there is a big catch! In order for that coordinate transformation to be well defined, $\theta$ had to be periodic with period $2\pi$. If not, then you would have a space where you took a sheet of paper, cut out a slice from it, and glued it back together, forming a cone. Therefore there is a singularity right at the tip, and we call this a "conical singularity."
Now going to the Euclidean black hole case, if you do a coordinate transformation $\rho^2 = 8M(r-2M)$ and expand near $\rho =0$ you get something like
$$ds^2 = d\rho^2 + \rho^2 \beta^2 d\tau^2+ \cdots$$
Therefore, in order for $\rho=0$ to remain merely a coordinate singularity the coordinate $\beta \tau$ must have periodicity $2\pi$ which means $\tau$ must have periodicity $2 \pi/\beta$. As a fun exercise determine $\beta$!