Let $\mathcal{A}$ be an L-structure with base set $A$ and $f$ a definable function. $$y=f(x)\Leftrightarrow \mathcal{A}\models \psi(x,y,\overline{a})$$ If we put another parameter $\overline{b}$ in place of $\overline{a}$ and we set $$B:=\{(x,y)\in A^2, \mathcal{A}\models\psi(x,y,\overline{b})\}$$ I think that the set $B$ could be empty. And in the case where it is not empty it does not need to define a function. What i mean is, it may exit to different $y_1$ and $y_2$ such that $(x,y_1)\in B$ and $(x,y_2)\in B$ .
My question: Is my understanding correct? and can we find a concrete example?
Yes. There are lots of examples. Take for instance $\mathcal{A}=(\mathbb{Z};+)$ and consider $$\psi(x,y,a)\equiv (a+a=a\wedge y=x+x)\vee (a+a\not=a\wedge \theta)$$ for any $\theta$ whatsoever. When we set the parameter $a$ to $0$ we'll get the function $x\mapsto 2x$, while any other choice of parameter will behave as horribly as you want depending on what $\theta$ is (e.g. take $\theta\equiv\top$ to get the total relation and take $\theta=\perp$ to get the empty relation, neither of which is a function defined on all of $\mathcal{A}$).