On the uniqueness of an analytic continuation for the Lacunary series $\sum_{n=0}^{\infty}q^{n^2}$

Question

Let $$f(q)=\sum_{n=0}^{\infty}q^{n^2}$$ For $|q|<1$. Then it can be expressed in terms of θ-functions: $$\theta_3(q)=\sum_{n=-\infty}^{\infty}q^{n^2}=2f(q)-1$$ That is also well-defined for $|q|<1$. But from here, formula (3.3.27) page 74: $$\theta_3^2(q)=1+4\sum_{n=1}^{\infty}\frac{q^n}{1+q^{2n}}$$ Which is well-defined over $\mathbb{C}\setminus\{q:|q|=1\}$ and creates an extension of $\theta_3^2(q)$. If we were to revert the square, then it would create branches, and can be chosen accordingly. Although the formula work just fine, it seems like the proof was nowhere to be found in the article (or I might have missed it due to my terrible English).

Question:

1. How to prove that $1+4\sum\limits_{n=1}^{\infty}\frac{q^n}{1+q^{2n}}$ matches $\theta_3^2(q)$ for $|q|<1$? If can, how does one possibly came up with this extension?

2. Can the extension be modified to define an unique analytic continuation of $f(q)$?