Your ULN2003A darlington darling is configured as an open collector output. Open collector outputs are great, because they let you connect lots of devices to a bus without loading it heavily. I2C, for example, uses open-collector outputs. Normally, the output pin would be connected to VCC with a pullup resistor. When you drive the input high, the BJTs begin conducting, resulting in a low impedance path to ground. The output is then low. When you release the input, the output pin has a high impedance to ground, and the line is pulled to VCC. When many open-collector devices are sharing a bus and one device outputs a logic low while the rest are outputting a logic high, the low wins out and pulls everything low without carrying all of the current from every connected logic-high device to ground.
To answer your questions, the reverse-biased diode on your VCC pin is definitely supposed to be there. It's used to clamp your output pin voltage to no higher than VCC plus the diode drop. That'd be so you don't fry anything else on the bus. That chip's full of protection diodes. Input to GND and Output to GND both have diodes that protect against a reversed polarity on the pins. Is this device meant to be hot-swapped? I expect this is all about shunting transient voltage spikes, like what you'd see from ESD or disconnecting a cable while current is flowing through it, to ground and protecting the device and everything upstream of it.
For your second question, I guess I jumped the gun and partially answered it at the top. Current shouldn't flow into the VCC pin at all. You're right in that, if the diode were reversed, you'd either have an output high or a hard short to ground. With the pull-up on the output pin, though, that's no longer the case.