Questions about the circuit of the Arduino Micro
When 5 V is available and Vin is not available


In the beginning only the body diode of T1 conducts. 
Hence, there will be (5V - diode drop) at the source of T1. 
that is only momentary,because, now we have a valid Source voltage. Gate is pulled low to ground via R9 (10kohm). 
Now, Since, Vgs is more negative, PMOS is ON. Once it is on, diode gets bypassed by the MOSFET. Hence, one will get almost 5V finally for the 3.3 V regulator . 
One can assume, the low on-resitance of PMOS will certainly create very
low ohmic drop.


When 5 V is available and Vin is also available   


The body diode may conduct in the beginning if 5 V is plugged in first.
The MOSFET too may turn on sooner and provide almost 5 V to 3.3 V regulator
As soon as Vin is connected (assuming that Vin is greater than 5 V), the MOSFET turns off because VGS is no more a negative value.
The current from USB will no longer be used because, the strong 5 V is available from 5 V regulator which is now powered from Vin. 


When 5 V is not available and Vin is available  


T1 will not be in the picture
T2 body diode can conduct and supply Vin to the U2 ( 5V regulator)
5 V regulator will feed the 3.3 V regulator as previous case


Function of D2  

There is a reset switch. Any spike arising out of the reset switch due to debounce can be sometimes higher than 5 V supply itself. This may inherently cause damage to the RESET pin of the MCU. Hence, placing the diode will start to conduct when voltage higher than 5.4 V (example) is present at the output of the switch. 

Consider it like a clamping circuit and clamping voltage is set to 5 V plus the forward voltage of the diode.  

Normally, there will also be diodes internal to the MCU pin for protection, but placing the external diode is definitely a good practice. The internal one is for emergency use only, i would say.