A.s == C (coulombs). It's a charge figure.
They shouldn't call it a power figure, since it has nothing to do with power, in and of itself. Can you ask the customer to clarify/correct, or is this something you have little recourse about?
If the supply voltage is 12V, you can use a maximum of Q/V = C (that's capacitance) or (10 A.s) / (12V) = 0.83 F (farad).
Seems like quite a lot, and even a modest motor would have trouble drawing quite that much inrush I think?
The time frame needs a definition (peak above what threshold?) but it further implies that the input current should be limited to, roughly, something around 20A. Depending on how that is defined (instantaneous, average over the duration, etc.?), it may be less stringent than that.
The energy associated with this charge, is also unclear; but if it is a linear capacitance being charged, then the energy is not the product again, but half. Consider E = 0.5 C V^2, then sub in C = Q/V. Mechanically speaking, it's the same relationship between momentum (kg.m/s) and energy (kg.m/s^2).

Tim