$7*8 = 56$. We imagine digits
0, 1, 2, 3, 4, 5, 6, 7, 8, 9clockwise (as magic numbers on Harry Potter's magic clock) and then $7$ times successively count $8$ numbers (clockwise, starting from $1$). After $7$th count we land exactly on $6$.
Credits and thanks to J.-E. Pin and Will Orrick from this site.
Besides, something very interesting was suggested by Bill Dubuque (see comments below), but I'm in the process of trying to approach and grasp it yet - it is very complicated (group theory/cyclic groups and star polygons / spirograph curves (roulette curves)).
But is there any reasonable child-friendly logic to quickly predict the tens digit (number of tens)? (without actual calculation)
As above, we can just count number of times we cross $0$ (zero) in the clock. This would give the tens digit.
Can we further simplify the way of finding number of crossing zero while doing those $7$ loops (counting $8$ successive numbers) on that Harry-Potter Clock? Number of crossing zero forms what sequence and how can it be easily constructed?
Besides, there is a method in Wikipedia (it helps with the last digit, but does not help with the tens digit and is much more complicated than Harry Potter Magic Clock above).
Maybe there is a simpler method - based on that we know the last digit (least significant digit) and both numbers being multiplied? (for a child it is difficult to count two things at once - number of transitions across zero and counting $8$ numbers in a row again and again in succession).
There are some patterns, but formulating an easy rule beats me.
Still, this is a very interesting academic question to me. Besides being a challenge to give full power of math (multiplication table) to kindergarten Einsteins at once to stimulate them (before just rot memorizing).
Below is the sequence of number of tens in multiplication table
(2: 2*1 ... 2*9, 3: 3*1 ... 3*9, etc)
2: (0), (0), (0), (0), (1), (1), (1), (1), (1), (2)
3: (0), (0), (0), (1), (1), (1), (2), (2), (2), (3)
4: (0), (0), (1), (1), (2), (2), (2), (3), (3), (4)
5: (0), (1), (1), (2), (2), (3), (3), (4), (4), (5)
6: (0), (1), (1), (2), (3), (3), (4), (4), (5), (6)
7: (0), (1), (2), (2), (3), (4), (4), (5), (6), (7)
8: (0), (1), (2), (3), (4), (4), (5), (6), (7), (8)
9: (0), (1), (2), (3), (4), (5), (6), (7), (8), (9)
Each Cell is digit of tens in the product of its coordinates
Here is the entire multiplication table (for reference).
As I noted in your prior question, we can intuitively represent such periodicity via star polygons (represented in toys like Spirograph). Since you are working $\!\bmod 10\,$ we use a "$10$ hour clock" of $10$ points placed equidistant along a circle. To get all multiples of $n$ modulo $10$ we start at $\,0\,$ then repeatedly add $\,n,\,$ by taking "big" steps of length $n$ along the circle. The path of this walk inscribes the $\{10/n\}$ star-polygon in the circle. Conveniently there is a nice YouTube video Star Polygons, by Linda Roper animating this case $\{10/n\},\,$ so please see there for further details on the basics (the image below is from there).
Let's construct the star polygon $\{10/6\}$ above. We obtain all multiples of $\:\!6\:\!$ by starting with $\:\!0\:\!$ then successively adding $\:\!6\pmod{\!10},\,$ yielding $\,0,\,6,\,12[\equiv\! 2],\, 8,\, 14[\equiv\! 4],\, 0.\,$ The $5$-point $\rm\color{#0ad}{aqua}$ star within $\{10/6\}$ is a graph of this process: starting at topmost point $(=0)$ draw an aqua line to the point $\:\!6,\,$ then draw a line from $\:\!6\:\!$ to $\:\!12[\equiv 2],\,$ etc. The resulting star polygon is the path traced by taking a walk on this $10$-point circle by taking steps of size $\:\!6.\,$ The journey visits the following points $\!\bmod 10\!:\ 6\Bbb Z = 6\Bbb Z + 10\Bbb Z = \gcd(6,10)\Bbb Z = 2\Bbb Z\,$ i.e. all multiples of $\:\!2,\:\!$ i.e. all evens. The $\rm\color{#f4a}{pink}$ star is the coset $\,1+6\Bbb Z = 1+2\Bbb Z = $ all odds, obtained by rotating the aqua star by one.
These star polygons can be plotted in a browser pane using the Geogebra app Circular Modular Addition and Multiplication Tables. To plot star polygon $\,\{m/n\}\,$ set modulo $ = m,\,$ plus $ = n$.
The Spirograph toy works the same way except it uses curves (vs. straight lines) to connect the successive points in the star polygon.
This (and related methods) provide great (visual) motivation for many results about cyclic groups - with the benefit they they can be understood long before one learns group theory. I've had success explaining such ideas to bright grade school students. It may prove helpful in your endeavor.
You can find an introduction to star polygons (and polytopes) in Coxeter's classic book Regular Polytopes. Below is an excerpt.