Complex-differentiability is about approximating a function by a complex-linear map. Real-differentiability is about approximations by a real-linear map. So, the question of when a real-differentiable function is complex-differentiable is reduced to the following linear algebra question:
Given two (say finite-dimensional) complex vector spaces $V,W$, and a mapping $T:V\to W$ which is $\Bbb{R}$-linear (for the underlying real vector space structure of $V,W$), what are the necessary and sufficient conditions to ensure $T$ is also $\Bbb{C}$-linear?
Well, for an $\Bbb{R}$-linear map to be $\Bbb{C}$-linear, it is necessary and sufficient that for all $v\in V$, we have $T(iv)=iT(v)$, in other words, $T$ needs to commute with multiplication by $i$.
In terms of complex structures, what does this mean? Well, $V$ is a complex vector space, so we can now define $J_V:V\to V$ as $J_V(v):=iv$, i.e $J_V$ is multiplication by $i$. Similarly, we can define $J_W$ on $W$. Note that both $J_V$ provides a complex structure on the underlying real vector space $V_{\Bbb{R}}$, and likewise for $(W_{\Bbb{R}},J_W)$. We can now answer the above question as follows: an $\Bbb{R}$-linear map $T:V_{\Bbb{R}}\to W_{\Bbb{R}}$ actually comes from a $\Bbb{C}$-linear map $T:V\to W$ if and only if
\begin{align}
T\circ J_V=J_W\circ T.
\end{align}
In the special case $V=W$, we can just write $J$ for the complex structure and succinctly write this equation as $[T,J]=0$.
By fixing bases, you can extract a corresponding matrix equation. Slightly more explicitly, let’s say $\dim_{\Bbb{C}}V=n,\dim_{\Bbb{C}}W=m$. Given a $\Bbb{C}$-ordered-basis $\beta_V=\{v_1,\dots, v_n\}$ for $V$, you get a corresponding $\Bbb{R}$-ordered-basis $\beta_{V_{\Bbb{R}},1}=\{v_1,\dots, v_n, iv_1,\dots, iv_n\}$ or another popular ordering choice is $\beta_{V_{\Bbb{R}},2}=\{v_1,iv_1,\dots, v_n,iv_n\}$. You can do a similar thing on $W$. Then, relative to these choices of bases, you get the corresponding matrix equation
\begin{align}
[T]\cdot [J_V]&=[J_W]\cdot [T].
\end{align}
Here, $[T]\in M_{2m\times 2n}(\Bbb{R})$, $[J_V]\in M_{2n\times 2n}(\Bbb{R})$ and $[J_W]\in M_{2m\times 2m}(\Bbb{R})$. Note that the choice of ordering of bases will give you slightly different matrix representations for the complex structures (see some remarks here).
In the case $n=m=1$, this just means you’ll have a single non-zero vector $v\in V$ which spans $V$. The matrix representation of the $\Bbb{R}$-linear map $J_V:V_{\Bbb{R}}\to V_{\Bbb{R}}$ relative to the ordered basis $\{v,iv\}$ is
\begin{align}
[J_V]&=
\begin{pmatrix}
0& -1\\
1&0
\end{pmatrix}=\Omega.
\end{align}
Similarly, we have $[J_W]=\Omega$. So, if we have the matrix-representation of $T:V_{\Bbb{R}}\to W_{\Bbb{R}}$ relative to $\{v,iv\}$ and $\{w,iw\}$ is $A=\begin{pmatrix}
a&b\\
c & d
\end{pmatrix}$, then we get that the condition for $T$ to be $\Bbb{C}$-linear is
\begin{align}
A\cdot\Omega=\Omega\cdot A,
\end{align}
i.e $[A,\Omega]=0$. If you work it out, this says the matrix $A$ must actually equal $\begin{pmatrix}a&-c\\c&a\end{pmatrix}$.
Granted, none of this was a geometric explanation, what I really want to stress is that complex-differentiability is by definition the requirement of having a complex-linear approximation. Everything else is literally just linear algebra to see what this condition of complex-linearity entails. In particular, the Cauchy-Riemann equations are also just a trivial consequence of complex-linearity. In typical presentations, people make it seem like Cauchy-Riemann equations are the ‘fundamental’ thing, but I think it’s much better to look at it the other way around; they’re just the algebraic conditions forced upon you once you write out the condition of complex-linearity explicitly in terms of real/imaginary parts (see the link above).
