I understand its something to do with the rank nullity theorem, but im not sure how they applied it to get the basis of the image. By my understanding, they took the leading entries of the rows of the reduced row echelon to get the kernel, so they took the leading entries of the columns to get the image? But if they did that, wouldn't the basis be (1,1,2,3) and (-1,3,4,7)?
In the reduced echelon form of the matrix there are leading $1$'s in the first and second column. This tells you that the first and second columns of the original matrix give a basis for the image of that matrix.
They did not derive the basis of the image from the kernel. The simple rule is that the columns which contain the pivots (or the leading $1$'s in each row) in the rref form are the first and the second columns. The rule for finding the basis of the column space is simply that you should choose the corresponding columns from the original matrix. In other words the first and the second column.