Do I have it right that the top circle in the 2nd-column to the right of the below packed square is $r(\sqrt{3}+1)$ units to the right of the left edge of the square. And by "is" I mean where the top circle in the 2nd-column kisses the top of the square:
Here, $r = 0.111382s$, where $s$ is the length of the square's side.
Yes, you can draw equilateral triangles between the centers of the circles. The first column is $r$ to the left of the right edge. The equilateral triangles have side $2r$ so height $r\sqrt 3$, making the second row of centers $r(1+\sqrt 3)$ in.
The above answer was based on the assumption that the circles are close packed, which results in the bounding box being a rectangle, not a square. The statement $r \approx 0.111382s$ comes from packomania where they are finding the largest circle for which you can fit (here) $20$ in a square. The bounding box for close packed circles is $(8+\frac 12\sqrt 3)r \approx 8.866r$ tall and $(2+4\sqrt 3)r \approx 8.928r$ wide. The sides are moved in a bit and the circles spread vertically to make the box a square. The side becomes about $8.978r$. You won't see the differences in wood. I wonder if there is a typo in the packomania figure. I would have expected the side to be between $8.866r$ and $8.928r$. We should be able to fit $20$ close packed in $8.928r$ with vertical room to spare.