Your writeup provides the background information that makes this a lot clearer.
Using the formula for E\_k at the start of section 7, we have
C\_k(x) = (1/2) \* ((1 + ix)^(2k) + (1 - ix)^(2k)) / (1 + x^(2))^k
and then by doing the rearrangement, and using C^' to denote the polynomials in your question because of Reddit, we get
4(C^(')\_k(x))^2 = 2(1 + x^(2))^k - (1 + ix)^(2k) - (1 - ix)^(2k).
Noticing that 1 + x^2 = (1 + ix)(1 - ix), the right hand side is seen to be -((1 + ix)^k - (1 - ix)^(k))^(2). Therefore
C^(')\_k(x) = (i/2)((1 + ix)^k - (1 - ix)^(k)).
The +/- ambiguity in taking the square root has been dealt with in this situation by desiring the first order term to be kx rather than -kx, but strictly speaking nothing in your definition stipulates this.
Anyway, with this expression in hand, we note that the main term is of the form a^k - b^k which, if m is a divisor of k, is divisible by a^m - b^(m). Therefore at the very least, if m divides k then C^(')\_m(x) divides C^(')\_k(x). For a further result, note that a^k - b^k = b^(k)((a/b)^k - 1) and now we have a product of cyclotomic polynomials in a/b. By distributing out the powers of b according to the degree of each term, we get a factorisation in terms of polynomials of a and b. Then plugging in a = 1 + ix and b = 1 - ix we get our nicer factorisation. I do not immediately see if the factors you get after the substitution are amenable to further factorisation.