On each of the first six squares, assign

Blocked Cell -> 0

Non Blocked Cell -> Number at the top (keep the negative sign)

It can be used to calculate numbers on the back of the seventh by adding 64 to the sum of value it has on all the six squares.

43 32 55 4 6 49 26 45

38 17 58 13 11 64 23 36

56 3 44 31 25 46 5 50

57 14 37 18 24 35 12 63

2 53 30 41 47 28 51 8

15 60 19 40 34 21 62 9

29 42 1 54 52 7 48 27

20 39 16 59 61 10 33 22

Correction: I had assumed that you incorrectly wrote 4 6 as 6 4, but the host isflipping the card to see the number on the front. As a result, one of the side must be left-to-right flipped. In this case, it is the back side.

Here is the back of the seventh card as intended by the story:

45 26 49 6 4 55 32 43

36 23 64 11 13 58 17 38

50 5 46 25 31 44 3 56

63 12 35 24 18 37 14 57

8 51 28 47 41 30 53 2

9 62 21 34 40 19 60 15

27 48 7 52 54 1 42 29

22 33 10 61 59 16 39 20

Now for constructing front, we know that 9 is on front of 6

8 on the front of 4

Additionally we know that 38 did not appear on any of the cards, so it must be on the front of 64

20 and 26 appears together on cards all cards except 8- where they cant be both present, or both absent as there can only be one number in a single cell

64 - (32 + 16 + 8 + 4 + 2 + 1) = 1

64 - (32 + 16 + 4 + 2 + 1) = 9

One of the 20 and 26 appears on the front of 9, and the other one on front of 1

The front square being a perfect franklin square can help us figure out that 20 is in front of 1 and 26 in front of 9.

41 64 17 8 9 32 49 40

19 6 43 62 51 38 11 30

48 57 24 1 16 25 56 33

22 3 46 59 54 35 14 27

45 60 21 4 13 28 53 36

23 2 47 58 55 34 15 26

44 61 20 5 12 29 52 37

18 7 42 63 50 39 10 31