Why does a micro-controller pin use push pull when configured as an output?
Push-pull means you can drive the output line high (by connecting it to VDD) or low (by connecting it to ground). A proper output needs to be able to do both of those things. [1]

If you're asking why the schematic shows transistors, that's a different question. The transistors form a CMOS inverter. They work exactly the same way as the triangle with a dot on the front. They're probably drawn separately because the transistors that drive the output pin are larger and can supply more current than the transistors used for internal logic. The two inverters together form the output buffer for the pin.

[1] You can also have an "open-drain" output that only has the pull-down transistor. You would then connect a pull-up resistor to the pin. The advantage of this is that you can make a "wired-OR" circuit where multiple open-drain outputs control the same line. The disadvantages are that the output's rise time is slower and the circuit uses extra power when the output is low.