The main difference between real analytic and complex analytic for functions in $\mathbb R^2 = \mathbb C$ is the following (I assume all the functions are smooth and we are working around $p=0$ for calculations):
Being real analytic at a point $p$ speaks about the growth of the derivatives in a neighbourhood of $p$. It is a very "analysis" property, because every smooth function $F : \mathbb R^2 \to \mathbb R^2$ can be "represented" in a neighbourhood of $p$ as
$$
F \equiv \sum_{n,m =0}^{\infty} \frac{\partial^{n+m}F}{\partial x^i\partial y^j}(0,0)\frac{x^i y^j}{i!j!}
$$
You can interpret $\equiv$ as you wish, but for example, by Taylor's theorem, these functions are equal to order $k$ for all $k \in \mathbb N$. The function is real analytic if $\equiv$ is an equality (this is, if the power series converges uniformly and coincides with $f$).
However, the obstruction to being complex analytic is bigger because it is "algebraic". Let me explain this:
Take $F$ as before. Because $\mathbb R^2 = \mathbb C$, you can think of $F=(F_1, F_2) = F_1+iF_2$. If $F$ is real analytic, then it is of the form
$$
F= \sum_{n,m=0}^{\infty} (a_{n,m}+ib_{n,m}) x^n y^m = \sum_{n,m=0}^{\infty} c_{n,m} x^n y^m
$$
For some complex $c_{n,m}$. You could define complex analytic functions as those that can be written in this form (convergent power series etc) but it doesn't seem very correct because $x,y$ are not the natural coordinates of the complex plane, so we use the formulas $z = x+iy, \bar z = x-iy$ to get $x = \frac{z+\bar z }{2}$ $y = \frac{z-\bar z}{2}$. After expanding all the powers, we get a representation
$$
F= \sum_{n,m=0}^\infty d_{n,m} z^n \bar z^m
$$
for some complex $d_{n,m}$. However, we have not done anything new. A funcion $F : \mathbb C \to \mathbb C$ can be written in this form iff it is real analyitic. Now, if you try to treat $\mathbb C$ as if it was a completely different field from $\mathbb R$, then complex analytic functions $F : F : \mathbb C \to \mathbb C$ should have a 1-dimensional power series representation (like real analytic functions $f : \mathbb R \to \mathbb R$), so you have to pick one variable. It turns out that the best thing is to pick $z$ as your variable. Then, complex analytic function should be of the form
$$
G = \sum_{n=0}^\infty e_n x^n
$$
Compare this to the representation of $F$. We are saying that a real analytic function is complex analytic when all the $d_{n,m}=0$ for all $m>0$. This is a VERY strong condition, and undoing the changes $z, \bar z$ yields it imposes an infinite set of algebraic conditions on the derivatives of $F$ wrt $x$ and $y$. We are saying that almost all the terms are $0$. It is basically reducing the information needed to encode a function from $\mathbb C^{\mathbb N \times \mathbb N}$ to $\mathbb C^{ \mathbb N}$ (I know these sets are equipotent, but think about it as linear vs quatratic growth).
The magic of complex analysis is being able to treat with this functions with so much ease, and this is a very very specific thing about the pair of fields $\mathbb R \subset \mathbb C$, and the main theorem is that if a $\mathcal C^1$ function $F : \mathbb R ^2 \to \mathbb R ^2$ satisfies the algebraic equations that I mentioned earlier for their first order derivatives (which are called the Cauchy-Riemann) equations, then it is complex analytic and therefore satisfies all these equations. This is one of the reasons why complex analysis is so beautiful and unexpected.
So, IMO the main you should keep is that $F$-analytic functions, where $F$ is a complete field like $\mathbb R $ or $\mathbb C$ (but also like the $p$-adics $\mathbb Q_p$) should be defined in terms of $F$ only, trying to generalize how you define them in $\mathbb R$ and that these definitions don't need to have anything to do with each other. For example, the difference between complex and real analytic is huge. However, just in the case $\mathbb R \subset \mathbb C$ it turns out that there is a very particular and interesting connection between the definitions.
