Yes, the triangle in the diode points in the direction of current flow, for "conventional" current, that is, current that flows from positive to negative. Yes, the diode blocks the discharge of the capacitor. Without it, the capacitor will discharge back into the load that you're monitoring, and it will discharge too quickly to sound the alarm.
Re connecting the battery, lamp, and power failure circuit, what is important is that the circuit is still connected across the lamp when you disconnect the battery.
Yes, I read your requirements, and they're the same as what I thought you wanted.
This is how the circuit works.
When the power supply is working normally, current flows through D1 and charges C1 up to about 0.7V less than the power source voltage (the 0.7V is lost in the diode junction). When power fails, C1 holds its charge because D1 prevents current flow from C1 back into the load that you're monitoring.
Q1 responds to the voltage difference between the base (on the left side in the diagram) and the emitter (at the top). The transistor will turn ON (conduct) when the base is brought negative relative to the emitter by about 0.7V. (Base negative with respect to emitter is correct for PNP transistors; it's positive for NPN transistors.)
When the circuit is connected to a power supply, let's say 12V, the positive input (top of diagram) has +12V on it, relative to ground (bottom of diagram). Therefore C1 will be charged to about +11.3V. The voltage on Q1 base will be pulled to +12V by R1. Therefore Q1 will be reverse-biased, i.e. the voltage on its base will be POSITIVE relative to its emitter, and it will be biased OFF. So no current flows through Q1.
When the power supply fails, the voltage on the circuit's positive input drops to zero, because the load pulls it to zero. R1 pulls Q1's base towards zero as well, but the resistance of R1 causes the current to be limited, and Q1's base voltage only falls low enough to turn Q1 ON (make Q1 conduct), i.e. Q1's base will be about 0.7V negative relative to its emitter. C1 keeps voltage present on the emitter.
When Q1 conducts, current flows from C1 into Q1 emitter, through Q1, out Q1 collector, and powers up the beeper circuit. The beeper circuit draws current, which will discharge C1 over a short period of time.
The transistor's base is not tied to V+. The EMITTER is tied to V+.
The BASE is the thick flat bit at the left in the drawing.
The EMITTER is the wire with the arrow. It points inwards towards the base for a PNP, and outwards for an NPN.
The COLLECTOR is the other terminal without an arrow on it.
