Performed the Danielson-Lanczos shuffling for FFT, but I don't know what to do next
I am not so sure about what you're asking for, but I will try to tell you something about this point of your question:


  The problem is that I'm a little confused on what to do with the resultant vector.


You have not a resultant vector, you have two vectors [0,2,4,6], [1,3,5,7] and by the D-L lemma you can compute the  DFT of the original vector as a linear combination of the DFTs of these two vectors.

Taking into account that the two shorter DFTs are periodic and that the combination coefficients for the second half are the negative of the coefficients used for the first half, we are lead either to a very simple recursive implementation of the FFT algorithm or to more complicated implementations that don't use recursion. Let's stay simple...

In a pythonesque pseudocode, the recursive implementation can be written as simply as

def fft(x,n,e):
    if n==1:
        X = x # the DFT of a scalar is the scalar itself
        return X
    else:
        X0, X1 = fft(x[0::2],n/2,e[0::2]), fft(x[1::2],n/2,e[0::2])
        X0, X1 = X0 + e*X1, X0-e*X1
        X = concatenate(X0,X1)
        return X


where 


x is the vector of which we want to compute the DFT
n is its length
e is a vector with the so called _twiddle_factors_, that is one half of the n-th roots of unity (from the lower part of the unit circle for forward DFT, from the upper otherwise)


and the [start:stop:step] notation stands for a selection of the elements of a vector, with a missing stop meaning to the end.

To call fft you have to provide the initial vector of twiddle factors

X = fft( x,  n, exp(-2*pi*1j*(0..n/2-1)/n)) # forward
x = fft(X/n, n, exp(+2*pi*1j*(0..n/2-1)/n)) # inverse


where 1j id the imaginary unit and (0..n/2-1) indicates an aritmetic progression from 0 to n/2-1, extremes included.