Euler Integral of a self-overlapping tube with a cusp singularity

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I am studying in depth the following paper on Euler calculus applied to target enumeration:

https://www.math.upenn.edu/~ghrist/preprints/eulerenumerationpart1.pdf

Within this paper there is an excerpt on the situation in which we have moving targets. I understand everything clearly except the following:

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I am confused about his integral computation above the figure. From my computation, $$ \chi(h >1) = \chi(\text{compact square}) + \chi(\text{self-overlap triangle with two points and an edge removed}) = (1-1+1) + (1-2+1) = 1. $$

Then $\chi(h > 0) = \chi(\text{entire figure projected to 2D}) = 0$.

Thus I still retrieve the same answer as the author in the total integral, but I am confused about his parenthetical computation involving the $1$'s.

In addition, what exactly does the author mean by "[It] is not a compact disc, but rather has a closed interval in the boundary removed," and "higher codimension piece of boundary."