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 :

$$f \; \text{ flat} \iff \nu :Y\to \mathbb N \text { constant}$$