Consider it as an array of arrays and this will work for sure.

int mat[][] = { {10, 20, 30, 40, 50, 60, 70, 80, 90},
                {15, 25, 35, 45},
                {27, 29, 37, 48},
                {32, 33, 39, 50, 51, 89},
              };


    for(int i=0; i<mat.length; i++) {
        for(int j=0; j<mat[i].length; j++) {
            System.out.println("Values at arr["+i+"]["+j+"] is "+mat[i][j]);
        }
    }

The most common way to do this is through for loops since they allow you to specify the index counters you need in a concise way:

for(int i = 0; i < scores.length; i++){
    for(int j = 0; j < scores[i].length; j++){
        scores[i][j] = in.nextDouble();
    }
}

If you specifically need to use while loops you can do pretty much the same thing, its just broken up into multiple lines:

int i = 0;
while(i < scores.length){
    int j = 0;
    while(j < scores[i].length){
        scores[i][j] = in.nextDouble();
        j++;
    }
    i++;
}

i and j are local to the for loops, you should declare i and j outside of the for loops at the top if you want to use them with a do while loop. This allows for a greater scope of the program

Regarding,

but the code gets stuck in a loop and I don't think it's doing what it's supposed to do.

Here:

for (int h = 0; h < correctedSums[1].length; i++) {

This loop will never end since h never changes within the loop. It should be:

for (int h = 0; h < correctedSums[1].length; h++) {

If you ran the code in a debugger, or used println's, you'd know what loop the code is stuck in, and this would allow you to inspect it immediately and correct it.

Assuming you only have to initialize the cells that get 1 value (the rest of the cells are 0 by default):

If a is much smaller than n, you can initialize the cells that get 1 value in O(n + a*n) time.

For example, if a == 0, all you need is to set the n cells of the main diagonal of the matrix ((0,0),(1,0),...,(n-1,n-1)) to 1.

If a == 1, you need to set the n cells of the main diagonal + the 2*(n-1) cells of the diagonal adjacent to the main diagonal.

...

If a = c, you need to set O(n) + O(2c*n) cells to 1, which is O(n + c*n).

To implement this algorithm, you'll need to replace your O(n^2) loop with 2*a+1 O(n) loops (one loop for each relevant diagonal).