An induction cooker works with a flat coil setup similar to the one shown below. The diameter of this coil depends on the exact model (in the range of 30 cm). This implies that the field is also extending quite a bit into the air above the cooker.
As the large field can be potentially hazardous (think metal ring on finger) there are relatively strict safety standards for induction cookers. The field of of the coil measured at a horizontal distance of 30 cm from the end of the coil and along the vertical axis (z=0 is the center of the coil) is shown below (the absolute values are not comparable with the field above the center but the profile is certainly comparable). The data is from this study: "B-Field Exposure From Induction Cooking Appliances" by C. Viellard and coworkers.
As you can see the field strength decreases by about an order of magnitude in a vertical distance of 200 mm. A very wobbly pan might be less effective but only if it is warped by more than a few cm.
I think there is some interesting physics to be had here. The rate of change of temperature depends on the rate of heat flow in from the electric heating element and the rate of heat flow out as heat is lost to the air. If we write the heat capacity of the hotplate as $C$ ($C$ is the traditional symbol for heat capacity) then:
$$ \frac{dT}{dt} = C \left( \frac{dH_{in}}{dt} - \frac{dH_{out}}{dt} \right) $$
where $dH_{in}/dt$ is the rate of heat flow in and $dH_{out}/dt$ is the rate of heat loss.
$dH_{in}/dt$ is simply the power being supplied to the hotplate. You can measure the current the hotplate draws from the mains and calculate the power that way, or you could simply use a power meter. The power in watts, call this $W$, is simply the energy in joules per second, so it's exactly what you need for $dH_{in}/dt$.
The rate of heat loss, $dH_{out}/dt$ is harder because it depends on how the hotplate is cooled. If the cooling is dominated by convention (it probably is) then the cooling will obey Newton's law of cooling and the heat loss will be given by:
$$ \frac{dH_{out}}{dt} = A \space (T - T_0) $$
where $T_0$ is the ambient temperature and $A$ is some constant to be determined experimentally. Put all this together and you'll get:
$$ \frac{dT}{dt} = C \left( W - A \space (T - T_0) \right) $$
You'll need to measure $C$ and $A$ experimentally. If you have a copy of excel to hand you can use its Solver to fit values of $C$ and $A$. Alternatively, you can get $C$ from the initial rate of temperature rise. When $T \approx T_0$ the heat loss is small and:
$$ \frac{dT}{dt} \approx CW $$
so if you know $W$ you can calculate C. You can calculate $A$ by heating the hotplate then turning the power off and letting it cool. As it cools the temperature variation is:
$$ \frac{dT}{dt} = -C A \space (T - T_0) $$
so if you know $C$ you can calculate $A$.
Best Answer
Your first assumption is not really correct. The induction cooker creates an electromagnetic field with a frequency in the range of 10 kHz to 100 kHz. The coil used in the cooker looks as follows:
The coil and the frequency are choosen in such a way that the field does not extend more than a few centimeters along the axis of the setup. Now when you put a ferromagnetic material on top it will rapidly get magnetized and demagnetized with with field created by the coil. Every ferromagnetic material exhibits hysteresis, which leads to dissipation. This dissipation heats the pot. This is not connected to the resistivity of the material, as eddy currents are not the main cause of the heating (a copper pot works poorly on an induction cooker).
With the field strength rapidly decreasing with distance, it is clear why the top of the pot and the lid do not become hot. There is no even current flow through the cookware.
To the second part of your question: Almost all electric energy is converted into heat, the total efficiency of input energy versus the heating of the food inside the pot is 84%, so very high compared to other methods but not perfect.
An electric shock is unlikely but not impossible. The Swiss health board recommends non-metallic spoons to avoid any current going through your body (Best practices, in German). Modern cookers have metallic or graphite surface to provide a path to ground for the pot which minimizes this issue.