Schottky diodes question. Which will power the board? Battery or USB?
That depends on the voltage on the battery.

If the battery voltage is greater than VBUS.5V plus a diode forward drop then D2 will be forward biased and D1 will be reverse biased. The battery will be driving the regulator and the USB will be effectively disconnected.

If the battery is less than VBUS.5V minus a diode forward the converse is true. D2 well be reverse biased and D1 will be forward biased. The USB will be driving and the battery will be effectively disconnected. However is will receive the diodes leakage current as a charge current.

Anywhere in between and the current will be divided between them in a ratio that depends on the forward voltage across their respective diodes.