We had a very large number of solutions come in for this challenge. Here are a selection for you to consider. 
Primary 6 at Victoria Primary School in Scotland sent in this very thorough solution

Firstly, we looked at the 4 numbers at the bottom and tried to work out how to get to 15.  As a class we agreed that this would take too long for us to try and work out  all the 4 digit numbers.  So next we worked out a system...

We started at the top and worked out all the 2 digit numbers that make 15 which were

14 + 1; 13 + 2; 12 + 3; 11 + 4; 10 + 5; 9 + 6; 8 + 7; 7 + 8; 6 + 9; 5 + 10; 4 + 11; 3 + 12; 1+ 14;

Then we looked at next row down and worked out that we couldn't have 1 + 14 and 14 + 1 because we we couldn't use 0 so nothing adds up to 0.  We took a pair of numbers each and worked out some of the group of 3 numbers which were:

And from that we were able to do the exact same to work out the four
numbers which are:

2221; 8111; 4122; 4211; 5212; 3211; 5211; 4122; 7112; 5114; 3213; 3123; 7112; 2131; 7112; 7223; 5211; 2214; 2311; 8231; 1312; 6113; 1127; 1118; 4212;

We really enjoyed the challenge and our (something? brains? hands? minds ?) hurts after it but in a good way.
We really needed to keep a growth mindset.
This has been explained very well but now have a look at the ones I've put in italics (and underlined) what do you think about them?


We also had a number of solutions sent in from  Peak School in  Hong Kong

Alyssa wrote:

I found that all of my solutions always had a 1 in the bottom row, and the
total of all the numbers in the bottom row was always an odd number.

Victoria K & F wrote:

These are our solutions:

Rules:
There has to always be a 1 in the bottom row.
In the bottom row will always be a number repeated twice.
The bottom row must equal an odd number.
The second to bottom row must equal 11, 12 or 13.
There must be a combination of odd and even numbers in every row.
 

Mathias wrote:

I found out 9 different ways to get 15 using different numbers.


 

Sophie and Dorika wrote:

We think that the bottom line has to add up to an odd number.
We also found out that there has to be a one in in the bottom row.
The middle line can't have a one in it.
The second to bottom row will equal to 11, 12, 13.
On the bottom row a number will always be repeated twice or three times.
There are only 4 pairs of numbers that can be used in the second row.

   
Extension:
We combined your idea and we did our own version using decimal points

                                              15
                                             9  6
                                           6  3  3
                                         4  2  1  2
                                    2.5 1.5 .5 .5 1.5
                             1.25 1.25 .25 .25 .25 1.25
                     .125 1.125 .125 .125 .125 .125 1.125


Chloé, Holly and Anna wrote:

We found these solutions by starting from the top row and working our way
to the bottom:

We found some rules after finding the solutions:

-The bottom row must equal an odd number (7, 9, 11)
-The second to bottom row equal 11,12 or 13
-On the bottom row there will always be a number repeated twice
-There has to be be one in the bottom row

Thanks for reading!

Jihun from  Dong Sung Elementary School in South Korea sent in the workthat had been done:

 

Well done all of you. It was very interesting reading about your methods and the things that  you found out that had to be. Keep up your good work.