Then draw four straight line segments, each from the center point to the perimeter, and each an integral number of centimeters long, so that if you wish to cut off an area of any integral multiple of 6 cm², you need only cut along two of these lines. Of course cutting off a multiple of 6 cm² will also leave a multiple of 6 cm².

What are the lengths of the four segments and how can they be arranged around the center?

I first started by finding what possible integral length cuts are possible from the center.  Obviously we have the 4 lines from the center to the centers of each of the sides, these are of length 3 and 6.  Now we also want it to contribute an integral area and thus we are only left with a cut from the center to a point 4cm from the center of the 12cm sides.

Thus I see 8 cuts for us to choose 4 from.

Now we need to be able to from 6cm^2 and thus need to pick a cut to the center of one of the 12cm sides and then a cut to a point 4cm either above or below this center.  Thus forming a 3,4,5 right triangle.

For simplicity let the vertices of the card be at
A: (0,0)  B: (6,0)  C: (6,12)  D: (0,12)
Then our first 2 cuts are to the points (0,6) and (0,10) and are of lengths 3 and 5 respectively.  These allow us to form areas of size 6,66,18, and 54.

Now we add cuts to (3,0) and (6,4) of lengths 6 and 5 respectively.

Now we have our four lines to cut along which are
X: (0,6)
Y: (0,10)
Z: (3,0)
W: (6,4)

And to form all areas that are multiples of 6 from 6 to 66 we need to make the following cuts
6: XY
12: ZW
18: XZ
24: YZ
30: XW
36: YW
42: XW
48: YZ
54: XZ
60: ZW
66: XY

And if you want 72 then obviously make no cuts as that is the entire card.

I am not certain but I think all other solutions are simply variations on this idea.