Sloping Binary Numbers

JavaScript (ES6), 53 bytes

n=>[...Array(n)].map(g=(j=1,i)=>j>i?0:j&i|g(j+j,i+1))

0-indexed. It's not often I get to use a recursive function as a parameter to map.

Mathematica, 46 bytes

Plus@@@Table[BitAnd[n+k,2^k],{n,0,#},{k,0,n}]&

Unnamed function taking a nonnegative integer # as input and returning the 0-index sequence up to the #th term. Constructs the sloping binary numbers using BitAnd (bitwise "and") with appropriate powers of 2.

Jelly, 11 bytes

ḤḶBUz0ŒDUḄḣ

Try it online!

Explanation

ḤḶBUz0ŒDUḄḣ    Main link. Argument: n
Ḥ              Double the argument. This ensures there are enough
               rows, since n + log2(n) <= 2n.
 Ḷ             Get range [0 .. 2n-1].
  B            Convert each number to binary.
   U           Reverse each list of digits. 
    z0         Transpose, padding with zeroes to a rectangle.
      ŒD       Get the diagonals of the rectangle, starting from the
               main diagonal. This gets the desired numbers, reversed,
               in binary, with some extras that'll get dropped.
        U      Reverse each diagonal.
         Ḅ     Convert each diagonal from binary to a number.
          ḣ    Take the first n numbers.

The transpose is the simplest way to pad the array for the diagonals builtin to work. Then the reverses are added to get everything in the correct order.