Understanding thermoelectric generator and Ohm's law
You can treat the TEG as an equivalent source of x volts with a series resistance of R.
According to the first chart, under realistic operating conditions, the open-circuit voltage will not exceed about 8.5V typically, well within the 18V recommended for the AMS1117.
If your objective is to extract the maximum output power, then pay attention to the last graph. All the situations will have that characteristic, just for different optimum loads.
How you use the information depends on what you are trying to calculate. If  you know the operating conditions then  you can figure out the optimum loading to get the most power out, the maximum voltage and the voltage at any current. Let's assume you have Th = 250°C and Tc = 50°C. You can just about get 3.3V out at optimum loading, which is about 2.75A, way more than that regulator can handle. The TEG appears as a 6.5V source (first graph) with less than 1.15 ohms in series (second graph). Remember that the 'matched' load resistance equals the source resistance for maximum power transfer.
With a 0.5A loading, the voltage should rise to 6V - 0.5*1.15 = 5.4V so your 3.3V linear regulator will be dissipating about 1W.
In practice it may be difficult to maintain those thermal conditions due to the large heat flow across the TEG, but that's outside the scope of this answer.
P.S. One would expect the heat flow to be slightly affected by the loading (more load current = more heat loss) as it acts like a TEC, but if we premise the calculation on the temperatures then that is not important. This is analogous to a generator that is harder to turn as the load current increases, but it's far more pronounced in a generator because of the roughly order of magnitude higher efficiency.