First, it must be shown that in order for such a chain to be formed, the numbers at the ends must be the same (by the way, thereby precluding the visible numbers we saw1 and 2from being the actual end numbers). Each number appears 8 times in the set of dominoes: once each in combination with zero through six except for twice on its appearance on the double (i.e., in combination with itself). In a chain of dominoes, the internal (touching) ends must contain an even number of each number, as adjoining ends match. That means the two end points must account for an even number of any number shown, and therefore must be two of the same number.
That shows that a matching number on the two tiles is necessary; is it sufficient? Yes. Call the number that matches between the two dominoes a, and the other number or numbers b (and possibly c). Then assign the remaining pip numbers to the other letters up to g and use one of these two layouts:
aa ac cb bd dc ce ed df fe eg gf fa ag gb bb be ee ea ad dd dg gg gc cc cf ff fb ba
or
ab bb bc cc cd dd de ee ef ff fg gg ga aa ad db be ec cf fd dg ge ea af fb bg gc ca
The way the two dominoes in the question were chosen, any of the 56 numbers (domino ends) in the set was equally likely to be chosen for each of the two visible positions. That includes each of the eight 1's and each of the eight 2's. The determination of which 1 or 2 is showing determines what the other number is on that domino. Note that the double1 for example has twice as many ways of appearing as the 1 in, say, 13 (but as many ways as 1 OR 3 in 13), because regardless of which way it's oriented, it is a possibility for what has happened in this instance.
So there are 8 possibilities for the first domino (including two that are double1), and 8 for the second. Ordinarily this would be 64 possible combinations, but the 12 domino cannot appear in both places, so there are really only 63 equally likely ways of the pair appearing as it did.
Out of these equally likely ways, which ones have one number from the first domino matching one number from the second, so that they can be placed on the ends of a chain? They are:
10; 20
11; 21
11; 21 (the other 1 being the one showing)
13; 23
14; 24
15; 25
16; 26
12; 20
12; 22
12; 22 (the other 2 being the one showing)
12; 23
12; 24
12; 25
12; 26
10; 21
13; 21
14; 21
15; 21
16; 21
This is 19 out of 63 possibilities, or 19/63 = .3015873015873...
A simulation confirms this:
RANDOMIZE TIMER
DIM pip(28, 2)
FOR low = 0 TO 6
FOR high = low TO 6
s = s + 1
pip(s, 1) = low: pip(s, 2) = high
PRINT s, low, high
NEXT
NEXT
DO
dom1 = INT(RND(1) * 28 + 1)
DO
dom2 = INT(RND(1) * 28 + 1)
LOOP UNTIL dom2 <> dom1
sid1 = INT(RND(1) * 2 + 1)
sid2 = INT(RND(1) * 2 + 1)
pip1 = pip(dom1, sid1): pip2 = pip(dom2, sid2)
IF pip1 <> pip2 THEN
tr = tr + 1
pip3 = pip(dom1, 3  sid1): pip4 = pip(dom2, 3  sid2)
IF pip1 = pip4 OR pip2 = pip3 OR pip3 = pip4 THEN
suc = suc + 1
END IF
PRINT USING "####### #.######## #.########"; tr; suc / tr; SQR((tr * suc / tr) * (1  suc / tr)) / tr
END IF
LOOP
Note that the simulation considers a valid trial any one in which the two showing numbers are unequal, as the probability just depends on the inequality of the two showing numbers, not on their identities.
When stopped after over 3,000,000 trials, the last line showed up to that point:
3378243 0.30136198 0.00024965
indicating 3,378,243 trials produced an average success rate of .30136198, and that if the theoretic success rate is within .00025 of that, that the results fall within one standard deviation of the expected results. The calculated probability and the simulation result do agree within that amount.
