How to prove Gauss's identities on the action of the operator $f' = x\frac{df}{dx}$ on Jacobi theta functions? 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.
 A: 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.
