In this post, the following two continued fractions discussed by Nicco are given,
$$A(q)= \left(\frac{\vartheta_2(0,q)}{2\,q^{1/4}}\right)^2= \cfrac{1}{1-q+\cfrac{q(1\color{red}-q)^2}{1-q^3+\cfrac{q^2(1\color{red}-q^2)^2}{1-q^5+\cfrac{q^3(1\color{red}-q^3)^2}{1-q^7+\ddots}}}}\tag1$$
$$B(q)=\frac{1}{2\,q^{1/2}}\frac{\vartheta_2(0,q^2)}{\vartheta_3(0,q^2)} =\cfrac{1}{1-q+\cfrac{q(1\color{blue}+q)^2}{1-q^3+\cfrac{q^2(1\color{blue}+q^2)^2}{1-q^5+\cfrac{q^3(1\color{blue}+q^3)^2}{1-q^7+\ddots}}}}\tag2$$
where $|q|<1$. One can see their beautiful affinity. On a hunch, I decided to investigate the similar cfrac,
$$C(q)=\cfrac{1}{1-q\color{red}-\cfrac{q(1\color{red}-q)^2}{1-q^3\color{red}-\cfrac{q^2(1\color{red}-q^2)^2}{1-q^5\color{red}-\cfrac{q^3(1\color{red}-q^3)^2}{1-q^7\color{red}-\ddots}}}}\tag3$$
Q1: Is it true that,
$$C(q) \overset{\color{red}?}= \frac{1}{q}\sum_{n=1}^\infty \frac{q^n}{1-q^{2n-1}} =\sum_{n=0}^\infty\sigma_0(2n+1)\,q^n = 1+2q+2q^2+2q^3+3q^4+\dots\tag4$$
where $\sigma_0(2n+1)$ is the divisor function? (I've checked it to hundreds of coefficients and it seems to hold. More info at A099774.) Note that the equality below is known,
$$\sum_{n=1}^\infty\sigma_0(n)\,q^n = \sum_{n=1}^\infty \frac{q^n}{1-q^n}$$
Q2: Can $C(q)$ be also expressed in terms of the Jacobi theta functions $\vartheta_k(0,q)$?
The answer to Q1 is yes. About Q2, I don't know yet.
As usual, I suppose that $q=\exp(2\pi\mathrm{i}\tau)$ for $\tau\in\mathbb{H}$ (complex upper half plane) and define $$q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$$ Thus we can consider Theta functions to be functions of $\tau$. I write $\vartheta_k(0\mid\tau)$ instead of $\vartheta_k(0,q_2)$. This is mostly to avoid multivalue issues with $\vartheta_2$.
Proposition: For $q,r\in\mathbb{C}$ with $|q|<1$ and $|r|<1$, we have $$\cfrac{1}{1-q-\cfrac{r\,(1-q)^2}{1-q^3-\cfrac{rq\,(1-q^2)^2} {1-q^5-\cfrac{rq^2(1-q^3)^2}{1-q^7-\cdots}}}} = \sum_{n=0}^\infty \frac{r^n}{1-q^{2n+1}}\tag{P}$$ Then the claim for your Q1 follows from setting $r=q$.
Furthermore, the already settled claim for $A(q)$ follows from setting $r=-q$ because there is the known series expansion $$\left(\frac{\vartheta_2(0\mid2\tau)}{2q_4}\right)^2 = \sum_{n=0}^\infty q^n \sum_{d\mid(2n+1)} (-1)^{(d-1)/2} = \sum_{n=0}^\infty \frac{(-q)^n}{1-q^{2n+1}}$$ where the latter Lambert-like series can be understood as a simple sieving machinery that generates an odd $d$ in every step and updates the counters associated with all odd multiples of $d$.
I suspect that there are simpler ways of proving proposition (P), but in the current series of posts regarding such continued fractions (1st, 2nd, 3rd, 4th), everything starts with the formula from Ramanujan's 2nd notebook, chapter 16, entry 11: $$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{*}$$ The original has the sign of $b$ flipped, but I like to emphasize the symmetry.
To obtain the continued fraction in (P), we would need $ab=-r$ and $a/b=-1$, which is fulfilled by $\{a,b\}=\{\pm\sqrt{r}\}$. However, this implies that $a+b=0$, and in that case, Ramanujan's formula gets reduced to $0=0$. We need to work around that.
We have already dealt with $a+b=0$, but only for $ab=q$, so we need a new twist here. First rewrite $$(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty = \overbrace{(-a;q)_\infty\,(a;q)_\infty}^{(a^2;q^2)_\infty} \left(\frac{(-b;q)_\infty}{(a;q)_\infty} -\frac{(b;q)_\infty}{(-a;q)_\infty}\right)$$ Then assume $0<|a|<1$ and recall the $q$-binomial theorem $$\frac{(-b;q)_\infty}{(a;q)_\infty} = \sum_{n=0}^\infty \frac{\bigl(-\frac{b}{a};q\bigr)_{n}}{(q;q)_{n}}\,a^{n}$$ which implies $$\begin{align} \frac{(-b;q)_\infty}{(a;q)_\infty}-\frac{(b;q)_\infty}{(-a;q)_\infty} &= \sum_{n=0}^\infty\frac{\bigl(-\frac{b}{a};q\bigr)_{n}}{(q;q)_{n}} \,(a^{n}-(-a)^n) \\ &= 2\sum_{n=0}^\infty\frac{\bigl(-\frac{b}{a};q\bigr)_{2n+1}}{(q;q)_{2n+1}} \,a^{2n+1} \\ &= 2(a+b) \sum_{n=0}^\infty\frac{\bigl(-\frac{b}{a}q;q\bigr)_{2n}}{(q;q)_{2n+1}} \,a^{2n} \end{align}$$ Voilà, we have a product of $(a+b)$ with a continuous function. Let's plug that into Ramanujan's formula $(*)$ and cancel the common factor $(a+b)$: $$\small\cfrac{1}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}} = \frac{2\,(a^2;q^2)_\infty \sum_{n=0}^\infty\frac{\left(-\frac{b}{a}q;q\right)_{2n}}{(q;q)_{2n+1}}\,a^{2n}} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty}$$ This is continuous across $-\frac{b}{a}=1$, so let us see what happens there. The denominator of the right-hand side reduces to $2\,(-a;q)_\infty\,(a;q)_\infty=2\,(a^2;q^2)_\infty$ and therefore gets cancelled with the $q$-Pochhammer symbol in the numerator. The series simplifies as $$\sum_{n=0}^\infty\frac{(q;q)_{2n}}{(q;q)_{2n+1}}\,a^{2n} = \sum_{n=0}^\infty\frac{a^{2n}}{1-q^{2n+1}}$$ Thus we obtain $$\cfrac{1}{1-q-\cfrac{a^2(1-q)^2}{1-q^3-\cfrac{a^2q\,(1-q^2)^2} {1-q^5-\cfrac{a^2q^2(1-q^3)^2}{1-q^7-\cdots}}}} = \sum_{n=0}^\infty\frac{a^{2n}}{1-q^{2n+1}}$$ and now we can replace $a^2$ with $r$ and are almost done.
We had to assume $0<|a|<1$ along the way, which covers all $0<|r|<1$, but the case $r=0$ needs to be checked separately. Now $r=0$ causes both sides of (P) to reduce to $1/(1-q)$ because the $r^0$ in the series is to be properly interpreted as $1$ by continuity. Q.e.d.