Is there any intuition or meaning regarding Cauchy-Riemann equations?

A matrix of the form $A=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ is a pure scaling and rotation (it does not scale different directions differently, nor does it have any "shear"). This is exactly the type of operation that is represented by multiplication by a complex number. Indeed you can identify $A \begin{bmatrix} x \\ y \end{bmatrix}$ with $(a+bi)(x+yi)$, using the usual identification $\begin{bmatrix} x \\ y \end{bmatrix} \leftrightarrow x+yi$.

What the Cauchy-Riemann equations say is that the Jacobian of the functions $u(x,y),v(x,y)$ is of this form (the diagonal elements are equal, the off-diagonal elements are exactly opposite) which means that it can be represented as multiplication by some complex number. This is needed in order to have the derivative of your complex function at a point be another complex number rather than a 2x2 matrix, as it would be for a generic differentiable $\mathbb{R}^2$-valued function of a $\mathbb{R}^2$ variable.

They're oddly absent from most treatments of complex analysis, but I like the Wirtinger derivatives, which also make Cauchy--Riemann transparent.

The idea is that any complex function $f(x+iy)$ of two real variables $x, y$ can be thought of instead as a function $f(z, \overline{z})$ where $z = x+iy$ and $\overline{z} = x-iy$. For example, the "modulus squared" function is $$f(x+iy) = |x+iy|^2 = x^2 + y^2 = z\overline{z}.$$

This substitution makes it natural to differentiate with respect to $z$ or $\overline{z}$ instead of $x$ or $y$. Formally, $$\begin{align*}\frac{\partial}{\partial z} &= \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right) \\ \frac{\partial}{\partial \overline{z}} &= \frac{1}{2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right).\end{align*}$$

As a sanity check, $\frac{\partial}{\partial z} z = 1$, $\frac{\partial}{\partial z} \overline{z} = 0$, $\frac{\partial}{\partial \overline{z}} z = 0$, $\frac{\partial}{\partial \overline{z}} \overline{z} = 1$. The usual product rule works for both of these derivatives, so for instance $$\frac{\partial}{\partial \overline{z}} z\overline{z} = \left(\frac{\partial}{\partial \overline{z}} z\right)\overline{z} + z \left(\frac{\partial}{\partial \overline{z}} \overline{z}\right) = z.$$

More generally, as far as polynomials in $z$ and $\overline{z}$ are concerned, we may treat $z$ and $\overline{z}$ as independent variables exactly as if we were applying $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ to a bivariate polynomial in independent variables $x$ and $y$.

The Cauchy--Riemann equations say precisely that $\frac{\partial}{\partial \overline{z}} f = 0$. The proof is easy: $$\begin{align*} 2\frac{\partial}{\partial \overline{z}} (u+iv) &= u_x + i u_y + iv_x - v_y \\ &= (u_x - v_y) + i(u_y + v_x) = 0+0i \\ &\Leftrightarrow u_x = v_y \text{ and } u_y = -v_x. \end{align*}$$

This relates very nicely to holomorphic functions as limits of polynomials. Namely, given a bivariate polynomial $\sum_{i, j=0}^N c_{i, j} z^i \overline{z}^j$, the derivative with respect to $\overline{z}$ is 0 if and only if $c_{0, j} = 0$ for all $j \geq 1$, i.e. if and only if it was a polynomial in $z$ after all.

In this sense, the Cauchy--Riemann equations are just saying we're restricting to limits of polynomials in $z$ and not polynomials in $z$ and $\overline{z}$.

A function $f:\Bbb C \to \Bbb C$ may be seen as a map $f:\Bbb R^2\to \Bbb R^2$. When differentiable, you can look at the total derivative $Df(x,y):\Bbb R^2\to \Bbb R^2$, which is a $\Bbb R$-linear map. But you started with a complex function, so $\Bbb R$-linearity won't cut it. You want to set back $\Bbb R^2 = \Bbb C$, $z=x+iy$ and say that $Df(z):\Bbb C \to \Bbb C$ is a $\Bbb C$-linear map. When can you say this? Precisely when the Cauchy-Riemann equations are satisfied. In which case $Df(z)$, as a linear map between $1$-dimensional (complex) vector spaces, is multiplication by a (complex) scalar. What is this scalar? $f'(z)$.

Bottom line: the Cauchy-Riemann equations measure the deviation from $Df(z)$ being $\Bbb C$-linear.