From what I know, the hyperbolic trigonometric functions are almost the same as the circular trigonometric functions ($\sin, \cos, \tan$, et cetera without the $h$ suffix), except they output when a line coming from the centre at the given angle hits the surface of a hyperbola, rather than a circle.
From what I've seen, the relevant hyperbola has asymptotes at $45^\circ$, $-135^\circ$, $225^\circ$, and $315^\circ$, so basically the lines $y = 0$ and $x = 0$, but rotated $45^\circ$. This means that a line told to go along any of these asymptotes — defined by the fact that the hyperbola will never meet them — surely will never meet the hyperbola because it basically is the asymptote!?
And anywhere that goes through, judging by the diagrams I've looked at on-line, the empty space above and below the origin (excluding where the hyperbola really is,) should, as well, surely be infinity, since it never gets to the hyperbola?
The summarised question: How come giving a hyperbolic function that treads along the asymptotes does not give infinity or some other undefined answer?
Because the argument to the hyperbolic trigonometric functions is not the angle that your line makes with the $x$-axis, but rather the area between your line, the $x$-axis and the hyperbola (although to be more exact, with the $a/2$ in the picture, it's the area between your line, its reflection across the $x$-axis, and the hyperbola, with a sign to differentiate between the two sides of the $x$-axis). See the image below:
Note that you could say that the regular sine and cosine functions do the same thing, taking in the area of a sector of the unit circle, so it is actually the same, in a way.