The USB wall charger is capable of providing up to 1.3A. Current draw is based on load, not on upon what the supply can provide. Obviously if a load requires more than a supply can provide, the load may not work correctly, or the power supply may fail (and cause damage if not protected).
A voltage meter places a high resistance across the test leads so that you can get an accurate voltage reading without drawing much current.
A current meter places a low resistance in series with the device under test so that you can get an accurate current measurement without adding a significant resistance. Your meter should have fuses for various current measurement ranges so that you don't damage the meter.
When you measure current provided by the wall charger, you should do so with something being charged. If you're just measuring current with your meter as the only connected item, you're effectively placing a short across the charger's terminals and it will therefore be unable to supply sufficient current.
When you connect your meter to the supply with any range that's under the 1.3A supply capability, you should theoretically blow the fuse in your meter for that range. I can't speak as to why the meter is showing unusual numbers for each range.
Connect a load to the charger, then insert the meter between the load and the charger to get a current measurement. Always start with the highest range and progress to lower ranges as applicable. Your meter measures 10A max in the 10 range, and the next lower range has a max of 200mA. Therefore, I would not measure current from your wall charger on any range other than 10 until you know for certain that the load is drawing less than 200mA - at which point you can safely switch to the 200mA range.
Note also that your meter indicates that the current measurement port is unfused and that it will tolerate a maximum value for only 10 seconds. Therefore, you might be getting unusual readings because the current measurement circuitry has been damaged. If you have only been very briefly measuring current with it, it might still be functional.
The real problem is not that you just want to measure current. If so, you'd put the multimeter in current mode and connect the leads. You want to measure maximum current a source can safely deliver. That's very different.
The correct answer is you can't measure this, you have to get it from the datasheet. That is because at best you can only determine what your single sample does, which does not tell you what restrictions you have to design to to guarantee safe operation over the variation of possible units. You also can't always know what "safe" is for the device just from the outside.
However, for something with a normal digital output, we can make a few assumptions. Again, these are assumptions only and without a proper spec we can't know whether we are damaging the device. Let's say that we assume that dragging the voltage down to 90% of its open-circuit level will not hurt the digital output. You didn't say what voltage this port puts out nor provide any link, so we can only use hypothetical values. If this is a 5 V output, for example, then that means we assume we can load it in the high state without damage so that the voltage goes down to 4.5 V.
To measure the current the port is sourcing in this case, you first have to find the load that drags the output voltage to 4.5 V. Hook up the current meter in series with a variable resistor to the port when it is driving high. Use a separate meter to monitor the port voltage. Keep increasing the load (decreasing the resistance) until the voltage goes down to 4.5 V. The current meter is then showing you the current that the port can source at what we assumed was the safe voltage of 4.5 V.
If you don't have a suitable variable resistor, you can use a succession of fixed resistors to get close enough. Most likely the port will be able to source a mA or a few mA. A 4.5 kΩ resistor would draw 1 mA at 4.5 V. Start with 10 kΩ and go down from there carefully. You will probably end up with a resistor in the 500 Ω to 5 kΩ range.
If you use fixed known resistors, you can dispense with the current meter altogether. Start with 10 kΩ and keep decreasing the resistance connected to the port until its voltage gets to 4.5 V. Since you know the voltage and the resistance, the current the port is sourcing is simply that voltage divided by that resistance.
Best Answer
This is a transformer - a CT with just a clamp (probably a bidirectional TVS) and no load resistor so you will need to supply a burden resistor in order to get a usable reading.
In the article you linked, 22\$\Omega\$ was used, and you can use something similar, such as 20.0\$\Omega\$ 1%. The output AC voltage will be proportional to the burden resistor up to some value. 20 ohms should be safe. You could also use the AC amperes range of your meter, provided you know that the input resistance is low enough and thus avoid using the resistor. Often AC voltage ranges are a bit more accurate.
The transformer will supply an output current which is related to the input current by the transformer turns ratio, provided the output voltage is not too high.