Multiplying the Schroedinger equation by $\psi^*$:
\begin{align}
\psi^*[i\hbar\frac{\partial\psi}{\partial t}] = \psi^*[-\frac{\hbar^2}{2m}\nabla^2 + V]\psi
\end{align}
Its complex conjugate by $\psi$:
\begin{align}
\psi[-i\hbar\frac{\partial\psi^{\ast}}{\partial t}]= \psi[-\frac{\hbar^2}{2m}\nabla^2 + V]\psi^*
\end{align}
and subtracting:
\begin{align}i\hbar[\psi^*\frac{\partial\psi}{\partial t} + \psi\frac{\partial\psi^{\ast}}{\partial t}] = \psi^*(-\frac{\hbar^2}{2m}\nabla^2\psi) + \psi^*V\psi\; -\; \psi(-\frac{\hbar^2}{2m}\nabla^2\psi^*) - \psi V\psi^*\end{align}
The potential terms cancel, and dividing by $\hbar$ and multiplying by $-i$:
\begin{align}[\psi^*\frac{\partial\psi}{\partial t} + \psi\frac{\partial\psi^{\ast}}{\partial t}] = i\frac{\hbar}{2m}[\psi^*\nabla^2\psi\; -\; \psi\nabla^2\psi^*]\end{align}
The left side is the derivative of the product $\psi^*\psi$, what is by definition the density of probability $\rho$:
\begin{align}\frac{\partial\rho}{\partial t} = \frac{\partial}{\partial t}[\psi^*\psi ] = i\frac{\hbar}{2m}[\psi^*\nabla^2\psi - \psi\nabla^2\psi^*]\end{align}
But now things are not so clear: creating the vectors $\nabla\psi$ and $\nabla\psi^*$, the right side can be expressed as the divergence of a vector:
\begin{align}\nabla\cdot[\psi^*\nabla\psi - \psi\nabla\psi^*] = \nabla\psi^*\cdot\nabla\psi + \psi^*\nabla^2\psi - \nabla\psi\cdot\nabla\psi^* - \psi\nabla^2\psi^* = \psi^*\nabla^2\psi - \psi\nabla^2\psi^*\end{align}
Of course if we call $-i\hbar/(2m)[\psi^*\nabla\psi - \psi\nabla\psi^*] = \mathbf{J}$, the expression becomes similar to the continuity equation:
\begin{align}\frac{\partial\rho}{\partial t} = i\frac{\hbar}{2m}\nabla\cdot[\psi^*\nabla\psi - \psi\nabla\psi^*] => \frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{J} = 0\end{align}
I understand the idea: if the probability decreases in any region, it must flow outside, so that the total probability is always one. But how such a flow relates to the vector difference in the brackets? I mean: the left side is a known subject, the square of the wave function is associated to the probability, but it is not the case of the right side.