For the above continued fraction we have (see paper [1] Proposition 2.9)
$$
\frac{2q^{1/2}}{1-q^2+}\frac{q^2(1-q^2)^2}{1-q^6+}\frac{q^4(1-q^4)^2}{1-q^{10}+}\frac{q^6(1-q^6)^2}{1-q^{14}+}\ldots=2q^{1/2}\sum^{\infty}_{n=0}\frac{q^{2n}}{1+q^{4n+2}}
$$
and ([1] relations (13),(27))
$$
\frac{k'_r}{1-k_r}=-1+\frac{2}{1-u_0(q^{1/2},q)}
$$
where
$$
\frac{u_0(q^{1/2},q)}{2q^{1/2}}=\frac{1}{1-q+}\frac{q(1+q)^2}{1-q^3+}\frac{q^2(1+q^2)^2}{1-q^5+}\frac{q^3(1+q^3)^2}{1-q^7+}\ldots
$$
Combining the above relations we get
$$
CF(q)=4q\frac{k'_r-k_r+1}{k'_r+k_r-1}\sum^{\infty}_{n=0}\frac{q^{2n}}{1+q^{4n+2}}=\frac{4q}{\sqrt{k_{4r}}}\sum^{\infty}_{n=0}\frac{q^{2n}}{1+q^{4n+2}}
$$
Hence we only have to show that
$$
\frac{4q^{1/2}}{\sqrt{k_r}}\sum^{\infty}_{n=0}\frac{q^n}{1+q^{2n+1}}=\theta_2(q)\theta_3(q)=\sqrt{\frac{2Kk_r}{\pi}}\sqrt{\frac{2K}{\pi}}
$$
or equivalently
$$
\frac{2q^{1/2}}{k_r}\sum^{\infty}_{n=0}\frac{q^n}{1+q^{2n+1}}=\frac{K}{\pi}
$$
But the Jacobi elliptic function $cn(u)$ has expansion (see [2] pg.55):
$$
cn(u)=\frac{2\pi}{Kk_r}\sum^{\infty}_{n=0}\frac{q^{n+1/2}\cos((2n+1)z)}{1+q^{2n+1}}\textrm{, where }u=\frac{2Kz}{\pi}
$$
and $cn(0)=1$ (see [2],pg.17), hence
$$
1=\frac{2\pi}{Kk_r}\sum^{\infty}_{n=0}\frac{q^{n+1/2}}{1+q^{2n+1}}
$$
By this last note the conjecture follows.
References
[1] N.D. Bagis and M.L. Glasser "Evaluations of a Continued Fraction of Ramanujan". Rend. Sem. Mat. Univ. Padova, Vol. 133, (2015)
[2] J.M. Armitage and W.F. Eberlein "Elliptic Functions". Cambridge University Press, (2006)
