P.444-445 of volume 3 of Gauss's Nachlass, in which Gauss wrote down an identity for the infinite series of $\vartheta^4_3(x)$ (this identity is the essence of Jacobi's four squares theorem), include several differential equations which, together with his identity for $\mathbb{log}(\vartheta^4_3(x))$ (which is dicussed here), apparently enabled him to derive the identities for $\vartheta^4_2(x),\vartheta^4_3(x),\vartheta^4_4(x)$. I believe the missing piece in the puzzle of answering this question is to understand the origin of these differential equations. I will describe them right now.
Denote $\vartheta_2(x),\vartheta_3(x),\vartheta_4(x)$ as $r,p,q$ respectively. Denote also $t=\frac{1}{p^2},u=\frac{1}{q^2}$. Then define the following mathematical operator: $t' = x\frac{dt} {dx}$ (although the $'$ symbol, this operator is not derivative). Then I would like to know on what grounds the following identities hold:
$$\frac{u}{t}-\frac{t}{u}=2(tu'-ut')=-4u^3t'' = 4t^3u''$$ $$\frac{t'''}{t''}+3\frac{t'}{t}=\sqrt{\frac{1}{t^4}+16\frac{t''}{t}}.$$ The significance of the first equation can be explained along the following line: $$\frac{u}{t}-\frac{t}{u}=2(tu'-ut')\implies p^4-q^4 = 4x(\frac{d\mathbb{log}(\frac{p}{q})}{dx})\implies r^4 = 4x(\frac{d\mathbb{log}(\frac{p}{q})}{dx})$$, so this identity connects $r^4$ with the logarithmic derivatives of $p,q$.
I will have much appreciation for an answer that will explain the principle behind such identities.
At the start let's use the standard notation regarding elliptic integrals and theta functions. All variables and functions in the answer are real.
Let $q\in(0,1)$ be the nome and let $k\in(0,1)$ be the corresponding elliptic modulus. The following equations are definitions \begin{align} \vartheta_2(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{1a}\\ \vartheta_3(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{1b}\\ \vartheta_4(q)&=\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}\tag{1c}\\ k&=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}\tag{2a}\\ k'&=\sqrt{1-k^2}\tag{2b}\\ K&=K(k)=\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{3a}\\ E&=E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{3b}\\ K'&=K(k')\tag{4a}\\ E'&=E(k')\tag{4b} \end{align} The following are then standard theorems based on above definitions \begin{align} q&=\exp(-\pi K'/K)\tag{5a}\\ k'&=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}\tag{5b}\\ K&=\frac{\pi}{2}\vartheta_3^2(q)\tag{5c}\\ \frac{dq}{dk}&=\frac{\pi^2q}{2kk'^2K^2}\tag{5d} \end{align} A key role is played by Legendre's identity $$KE'+K'E-KK'=\frac{\pi}{2}\tag{6}$$ in the proof of above theorems. The following identity $$\vartheta_3(e^{-\pi/ s})=\sqrt{s}\vartheta_3(e^{-\pi s}),s>0\tag{7}$$ is another essential ingredient used in the proof.
Now we switch to the notation used in question. The symbols $k,K$ have same meaning as in previous part of the answer and instead of $k',K'$ we use variables $l,L$ so that $l=\sqrt{1-k^2}$. The nome $q$ is now replaced by $x$.
The identity $(5d)$ is crucial here and we rewrite it using the new notation as $$\frac{dx}{dk}=\frac{\pi^2x}{2kl^2K^2}\tag{8}$$ We can rewrite $(5b),(5c)$ as $$\frac{p^2}{q^2}=\frac{1}{l},K=\frac{\pi}{2}p^2\tag{9}$$ Taking logs and differentiating with respect to $k$ we get $$2\frac{d}{dk}\log(p/q)=\frac{k}{l^2}$$ Using $(8)$ we get $$x\frac{d}{dx}\log(p/q)=\frac{k^2K^2}{\pi^2}$$ Using second equation from $(9)$ and definition $(2a)$ we get $$4x\frac{d}{dx}\log(p/q)=r^4$$ which is essentially the first equality in question.
The equation $(5d)$ can also be seen as an identity for $p^4,q^4$. We can rewrite it in the form $$q^4=4x\frac{d}{dx}\log (r/p)$$ and using it one can obtain the Lambert series for $p^4,q^4$.
The second equality in question involves second order derivatives and I think it is equivalent to the second order differential equation satisfied by complete elliptic integrals $K, L$ of first kind.