Define f by sending the left coset xH of H in G to the element xgx^(-1). We need to show that f is well defined: if xH and yH are the same coset of H in G, then from xH = yH we deduce y^{-1} x H = H and hence that y^{-1}x is in H. This means by definition of H that (y^(-1) x)^(-1) g (y^(-1) x) = g, and hence that x^(-1)ygy^(-1)x = g. If you multiply both sides of this equation on the left by x, and both sides of it on the right by x^{-1}, you see ygy^{-1} = xgx^{-1}, which shows that our rule is well defined.
We claim that f is one-to-one: suppose that xH and yH are cosets of H in G and that f(xH) = f(yH), ie, that xgx^{-1} = ygy^{-1}. Multiplying both sides on the left by x^{-1} and on the right by x we deduce g = x^{-1} y g y^{-1} x, and hence that g = (y^(-1) x)^(-1) g (y^(-1) x), and hence by definition of H that y^{-1} x is in H. But then the coset yH contains the element y*y^{-1} x = x, and since distinct cosets cannot have a nonempty intersection, we must have that xH = yH. This proves that f is one-to-one.
We claim that f is onto. If y is in K(g), by definition of K(g) there is z in G with y = zgz^(-1), but then by definition of f we have that f(zH) = zgz^{-1} y. This shows that f is onto.
Since f is a bijection from the set of left cosets of H in G to K(g), we conclude that |K(g)| = [G:H] and hence that |K(g)| divides |G|.