+-----+-----+-----+
|     |     |     |
|  1  |  2  |  x  |
|     |     |     |
+-----+-----+-----+
|     |     |     |
|  4  |  6  |  y  |
|     |     |     |
+-----+-----+-----+

With the numbers chosen, four of the values could be obtained in two different ways each, and the rest could be obtained in only one way each.

Your job is to figure out the numeric values of x and y.

Note that the square portion already given, containing 1, 2, 4 and 6, would, by itself satisfy the mentioned criterion: all the values from 1 to 13 can be achieved as the total of orthogonally connected stamps, such as 1, 2 and 4 to make 7 (the 6 and the 1 are not orthogonally connected, so that would not be allowed).

Ady, my solution was both partially analytical and computer aided. 
Given that there were four values that were repeated exactly twice, I knew that at least one of the numbers needed to be equal to a combination of the the other four distinct numbers and that that new number added to one of the existing four numbers should total the value of a combination of the remaining numbers. As the number 1 and 4 are not adjacent to x or y, the combination value to be added could not be 1 or 4. Thus, the upperbound value of the smaller possible value of x or y was 12, with 12 and 13 being two of the repeated values:
(6+4+2 ; 12), (6+4+2+1 ; 13) - and a lowerbound value of 7 for x and 9 and 10 being two of the repeated values: (6+2+1 ; 7+2), (6+4 ; 7+2+1). In like manner, the other number, x or y, would need be equal to a combination of the other five distinct numbers, keeping in mind a need for four combinations needed, an upperbound then would be 25 (13+6+4+2).
One could analytically hone the possibilities down to the solution,
but as I sometimes don't wish to go through all the steps manually, I do get the help from a computer program or application such as a spreadsheet - as I did in this case.