the step in the gate voltage is due to the miller-effect talked about earlier in this post.
This is a common mistake in driving MOSfets, you do not calculate with the Cgs capacitance to determine the switching time (and so current needed). The Cgd contributs more to the actual switching time than the Cgs.
Switch on cycle :
1. When the gate driver switched on, you start charging the Cgs capacitance, nothing happens yet to the Vds until the Vth of the FET is reached.
2. When the Vds of the MOSfet starts to change (the beginning of the switching), you need to change the voltage on the Cgd capacitance from about the Vds to -Vth. This because Vds = Vdg + Vgs (the voltage on Cgs + the voltage on Cgd = Vds)
The rate Vds going down ONLY depends on how fast you can discharge the Cgd-capacitor! This energy (current) comes entirely from your gate drive circuit, so the more current you supply to the gate, the faster Cgd discharges, the faster Vds goes down, and the faster the switching takes place.
3. When the switching is over (Vds ~ 0V), you only charge Cgs to a higher voltage.
That's why the manufacturers of MOSFets give curves with 'total gate charge', and this on different Vds levels, because the importand parameters are Vds and Cgd. Since Cgd is voltage depended on its own, it is generally not enough to use only a single capacitance value.
So you can compare the switching time with the duration of the plateau on the gate voltage. Doubling the current, will double the switching speed.