I do not like the statement of the question, because you are identifying the hyperbolic plane with the Poincaré disk model (apparently -- nothing is said about the model, so it could be actually the Beltrami-Klein model too). In my experience, the Minkowski hiperboloid model is much more "natural" and much better for working with hyperbolic geometry, and the Poincaré disk model is great only for visualizations and some specific arguments.

Anyway, some intuition why the sum in question should be infinite. I find this intuition convincing, but I guess it would take some work to turn this into a real proof.

Consider the upper halfplane model instead, and only the points with $x \in [0,1]$ and $y \leq 1$. We will show that the sum of $y$ coordinates of all the grid points in this sector is infinite. This will also show that your sum in Poincaré disk model is infinite too, because the Poincaré disk model behaves the same as the halfplane model in the limit (like this).

Isometries of the halfplane model include horizontal translations $(x,y) \mapsto (x+a,y)$ and scaling $(x,y) \mapsto (ax,ay)$. Take $a<1$. Since regular hyperbolic tilings have constant density, there are as many grid points in the rectangle $[0,1] \times [a^{k+1},a^k]$ as in $[0,a] \times [a^{k+2}, a^{k+1}]$ (by scaling isometry). Therefore, there will be $1/a$ as many points in the rectangle $[0,1] \times [a^{k+2}, a^{k+1}]$ (by translation isometry). Hence, the number of points in this rectangle is asymptotically $\Theta(1/a^k)$. The sum of $y$ coordinates of all points in the rectangle $[0,1] \times [a^{k+1}, a^k]$ will be $\Theta(a^k)$ ($y$ coordinate) times $\Theta(1/a^k)$, which is $\Theta(1)$. Since our region is the sum of $[0,1] \times [a^{k+1}, a^k]$ for all $k \geq 0$, the sum in question is $\sum_{k=0}^\infty \Theta(1) = \infty$.