MOSFET Driver operation question
You've read the "Wide Input Supply Voltage Operating Range" incorrectly. The range is from +4.5V to +18V for normal operations.
"Absolute Maximum Ratings" refers to what voltages will not damage the device. "Input Voltage" is referring to the voltage at the Input pin, not the main power input VDD. It is saying that even accidentally applying negative 5V, up to 0.3V above your main supply voltage (VDD), to the Input pin will not hurt the device.
When chips label multiple pins as the same name, they always mean they are internally electrically connected, unless they very explicitly state otherwise. In this case, they have provided two terminals to get power into the device because the losses associated with using only a single pin would be higher. They have also provided two pins for the output current for the same reason. In general, always connect all internally connected pins to each other externally. However, if you need to cut corners for other reasons, things generally still work if you don't connect all of the pins.
Yes, because the rise &amp; fall times are 25ns, if you used an operative frequency of 20MHz (a period of 50ns) and 50% duty cycle, the output FETs would basically never turn on. You would just dump a bunch of energy (aka heat) into the FET gates and the load would basically never turn on. Running at 14Mhz will get very close to this and you might overheat and blow up your FETs.
Output will be tied to either VDD or GND, depending on the state of Input. Therefore, when Input is at 5V, relative to GND, Output will be connected to VDD.