How can TMS induce enough depolarization to cause an action potential?
I think my mistake was to believe that the electric field has to be somewhat perpendicular to the membrane, so that it can directly apply a voltage difference across the membrane, and that the resistance inside and outside the axon was the same, so that a tangential field was useless.

In fact these are false : depolarization and action potential creation happen where the axon is parallel to the electric field and where the gradient of the tangent field is the bigger.

Here you have a complete explanation but I think we can make a simplified one :
http://stbb.nichd.nih.gov/pdf/Nerve_Fiber_Stimulation_Model.pdf

Let's look at a membrane model (Hodgkin-Huxley) : 



As we said if we try to apply an electric field $\vec{E}$ along $\vec{y}$ , it should be incredibly strong to produce a trans-membrane potential $V$ strong enough to create an action potential.

In between the inside an the outside of the membrane you have all the channels and the capacitance of the membrane. but I think the crux of the matter here are the resistances inside and outside : the small surface of a section of axon limits the conductivity, whereas the outside is supposed to be bigger so that its resistance is negligible.

As a result if you apply a gradient $\frac{dE_{x}}{dx}$ you can create trans-membrane current and potential:
because of the gradient, the fields $E_{1} = E(x)$ and $E_{2} = E(x+\Delta x)$ are different : $E_{2} = E_{1} + \frac{dE_{x}}{dx}\Delta x$.
Then using electricity laws : 
$$i_{m} = I_{1} - I_{2}$$
$$i_{1} = -\frac{ E_{1} \Delta x}{R}$$
$$i_{2} = -\frac{ E_{2} \Delta x}{R}$$

then :

$$i_{m} = \frac{\Delta x}{R} (E_{2}-E_{1}) = \frac{\Delta^{2} x}{R} \frac{dE_{x}}{dx}$$

This trans-membrane current will produce a potential across the membrane which will be proportional to $\frac{dE_{x}}{dx}$, electromagnetic stimulations shows that TMS can easily produce gradients strong enough to create actions potential.