As the other answer pointed out, with the typo fixed, the equation is $$\frac{4}{R}\prod_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$ After replacing $R$ with $q^{-1/2}$ we get $$4q^{1/2}\prod_{n=1}^{\infty} \left(\frac{1+q^{2n}}{1+q^{2n-1}}\right)^4 = 4q^{1/2}\left(1+ \sum_{n=1}^ {\infty} q^{n(n+1)}\right)^2 \Big{/} \left(1+2 \sum_{n=1}^{\infty}q^{n^{2} }\right)^2.$$ The left side is $k(q)$ and the right side is $\theta_2(0,q)^2/\theta_3(0,q)^2$ which is a standard equation. See, for instance, DLMF equation 20.9.1.

As the other answer pointed out, with the typo fixed, the equation is $$\frac{4}{R}\prod_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$ After replacing $R$ with $q^{-1/2}$ we get $$4q^{1/2}\prod_{n=1}^{\infty} \left(\frac{1+q^{2n}}{1+q^{2n-1}}\right)^4 = 4q^{1/2}\left(1+ \sum_{n=1}^ {\infty} q^{n(n+1)}\right)^2 \Big{/} \left(1+2 \sum_{n=1}^{\infty}q^{n^{2} }\right)^2.$$ The left side is $k(q)$ and the right side is $\theta_2(0,q)^2/\theta_3(0,q)^2$ which is a standard equation. See, for instance, DLMF equation 20.9.1 and Abramowitz and Stegun equations 16.38.5 and 16.38.7.

As the other answer pointed out, with the typo fixed, the equation is $$\frac{4}{R}\prod_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$ After replacing $R$ with $q^{-1/2}$ we get $$4q^{1/2}\prod_{n=1}^{\infty} \left(\frac{1+q^{2n}}{1+q^{2n-1}}\right)^4 = 4q^{1/2}\left(1+ \sum_{n=1}^ {\infty} q^{n(n+1)}\right)^2 \Big{/} \left(1+2 \sum_{n=1}^{\infty}q^{n^{2} }\right)^2.$$ The left side is $k(q)$ and the right side is $\theta_2(0,q)^2/\theta_3(0,q)^2$ which is a standard equation. See, for instance, DLMF section 20.9.1.

As the other answer pointed out, with the typo fixed, the equation is $$\frac{4}{R}\prod_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$ After replacing $R$ with $q^{-1/2}$ we get $$4q^{1/2}\prod_{n=1}^{\infty} \left(\frac{1+q^{2n}}{1+q^{2n-1}}\right)^4 = 4q^{1/2}\left(1+ \sum_{n=1}^ {\infty} q^{n(n+1)}\right)^2 \Big{/} \left(1+2 \sum_{n=1}^{\infty}q^{n^{2} }\right)^2.$$ The left side is $k(q)$ and the right side is $\theta_2(0,q)^2/\theta_3(0,q)^2$ which is a standard equation. See, for instance, DLMF equation 20.9.1.