Instead of trying to say what flatness in analytic geometry means I'll give you some street-fighting tricks for recognizing whether a morphism of analytic spaces ( not necessarily reduced) $f:X\to Y $ is , or has a chance to be, flat.
a) A flat map is always open.For example the embedding $\lbrace 0\rbrace \hookrightarrow \mathbb C$ is not flat.More generally, the embedding of a closed subspace $X \hookrightarrow Y$ is never flat.
b) An open map need not be flat: think of the map from the double point $Spec(\mathbb C[\epsilon ]/(\epsilon^2))\to Spec(\mathbb C)$ from the double point to the reduced point single point.
c) Given the morphism $f:X\to Y $ , consider the following property:
$\forall x \in X, \quad dim_x(X)=dim_{f(x)} (Y) +dim_x(f^{-1}(f(x)))$ $\quad (DIM) $
We then have: $ f \; \text {flat} \Rightarrow f \;\text { satisfies } (DIM)$
For example a (non-trivial) blowup is not flat.
d) For a morphism $f:X\to Y $ between connected holomorphic manifolds we have:
$$f \text { is flat} \iff f \text { is open} \quad \iff (DIM) \;\text {holds}$$ For example a submersion is flat, since it is open.
e) Given two complex spaces $X,Y$ the projection $X\times Y\to X$ is flat ( Not trivial: recall that open doesn't imply flat!).
f) flatness is preserved by base change.
g) The normalization $f:X^{nor} \to X $ of a non-normal space is never flat.
For example if $X\subset \mathbb C^2$ is the cusp $y^2=x^3$, the normalization morphism $\mathbb C\to X: t\mapsto (t^2,t^3)$ is not flat.
h) Given a finite morphism $f:X\to Y $, each $y\in Y$ has a fiber $X(y) \subset X$ and for $x\in X$ we can define $\mu (x)= dim_{\mathbb C} (\mathbb C \otimes_{\mathcal O_{Y,y}} \mathcal O_{X,x})$ . Now for $ y\in Y$ we put $\; \nu (y)= \Sigma_{x\in X(y)} \mu(x)$ and we obtain :
$$f \; \text{ flat} \iff \nu :Y\to \mathbb N \text { constant}$$