In the diagram you posted, the “vertical” 6-2 domino that you outlined can’t represent the 6-2 domino because you have already allocated the “6” to the 6-6 domino above and there is only one 6-6 location in the diagram.
This means that the horizontal 2-6 domino that you have highlighted with a question mark has got to be the 6-2 domino (or the 2-6 domino, they are the same domino). There are no other pairs of 6-2 in the diagram.
Yes, the 6-4 domino can be a 6-4 or a 4-6 depending on its orientation. However, in the diagram, that particular domino has been laid N-S with the six dots at the N end and the four dots at the South end.
The 6-2 domino that you highlighted with ? mark has been laid E-W with the six dots at the E side and the two dots at the W side.
In a standard box of 28 dominoes, there is only one of each denomination starting at 6-6 and working down to 0-0. Each domino is represented in the diagram. Each appears only once.
Each domino is a 2x1 rectangular tile. It is divided into two halves, each a 1x1 square. Each square contains 0 to 6 dots. The tile with the most dots has 6 dots in each square, ie 12 dots in total. You have identified where this tile is placed in the diagram. You have also identified the 6-2 domino (by eliminating one of two possibilities because of the position of the 6-6 domino.) I marked the 6-4 domino in the starter diagram.
You are correct. There is a systematic way of solving the puzzle.
I started by looking for unique pairs of adjacent numbers, eg 6-6. There is only one such pair, ie at the top. This pair therefore must form the 6-6 domino. I drew a rectangle around it and ticked off the 6-6 domino in my standard list.
I then looked for pairs of 5-5. But there was more than one such possible sets. So at this stage I was unable to decide which of those pairs formed the 5-5 domino. And I moved onto the 4-4 search.
I worked my way through all 28 standard dominoes, looking for unique pairs and I think I found about half a dozen. But it was a good start.
I then noticed that a few numbers would be left “orphaned” ie isolated, without an adjoining square unless they were partnered with a specific adjoining square. This meant I could eliminate all other pairs of that combination and be certain that I had found a domino, outlined it as such, ticked it off my list, and moved on.
It took a bit of careful thinking, but it all worked out in a satisfyingly way !