Why is there a spike in voltage over an inductor double the peak voltage of a square wave input?
It is useful to remember the inductor constitutive relationship: $$v_L(t)= L \frac{di_L(t)}{dt} $$
That is, the voltage across an inductor is proportional to the derivative of the current through it.

When you apply a constant voltage of, say, 1V at the input node (I assume your circuit resembles that of the schematic below) there will be a constant current flowing on the resistor, so the inductor will behave as a short-circuit, thus presenting a zero voltage drop (look at your plot: the red curve actually tends to zero!!!). 







simulate this circuit &ndash; Schematic created using CircuitLab

When the voltage at the input drops, inductor current continuity imposes the same current to flow as before,so the same voltage drop across the resistor -as before. Hence the additional negative voltage across the inductor.
Similar considerations also hold for the positive step. 

One may wooden why steps in inductor current cannot occur. 
Actually, when inductor current steps show up, something different will occur before the voltage goes to infinity, for example a spark when you open a switch lacking the suitable protection (normally provided by a freewheeling diode in power applications). 



From a mathematical standpoint, it is physically feasible to force a voltage (not current!!!) step across an inductor and a current (not voltage!!!) step across a capacitor; BEWARE: I am talking about steps applied to the inductor (or capacitor), NOT to the overall circuit. 
If you have some knowledge of control system theory, this is dubbed as the principle of integral causality. But I don't want to go deeper in that.
I hope this helps.