Consider the Maxwell equation equivalent to Faraday's law:
\begin{equation}
\boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}
\end{equation}
If you perform a surface integral of the terms on both sides you get:
\begin{equation}
\begin{split}
&\iint _{S} (\boldsymbol{\nabla} \times \mathbf{E}) \cdot \mathbf{dS} = \oint _{\partial S} \mathbf{E} \cdot \mathbf{dl} \\
-&\iint _{S} \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{dS} = -\frac{\partial}{\partial t} \iint _{S} \mathbf{B} \cdot \mathbf{dS} = -\frac{\partial \Phi}{\partial t}
\end{split}
\end{equation}
The electric field $\mathbf{E}$ will have the same direction as the current density $\mathbf{J}$ and thus the "direction" of the electric current $I$ on a 1-dimensional path (essentially either clockwise or counter-clockwise on the coil).
If you move the north pole of the magnet towards the coil or the south pole away, the direction of the magnetic field is the same as the normal vector of the surface of which the solenoid coils (i.e. your 1-dimensional paths), meaning the inner product is positive and the change in magnetic flux is also positive. Due to Faraday's law, the integral of the electric field on the coil must be negative, meaning the electric field (and thus the current) are in the opposite direction of the one you chose as positive for the coil i.e. clockwise. Hence for north pole approaching or a south pole moving away, the current runs counter-clockwise.
When the north pole is moving further away or the south pole is moving closer, the opposite is true: the change in flux is negative, thus the line integral of the electric field is positive. By the same logic as above, the current then runs clockwise.
