I am referring to the half wave rectifier from the app note
https://ww1.microchip.com/downloads/en/appnotes/01353a.pdf
While the positive half is understandable, the negative half puts the op-amp output to a large negative voltage theoretically, and the diode D2 'shorts' that voltage by supplying lot of current thru ground?
What am I missing regarding the circuit's working?
I don't think you are missing anything. If the op-amp has 0V as Vee then it will do nothing (but the input going below Vee will likely cause problems). If the op-amp has a negative supply then it will short the (railed) output through D2 (i.e. output will be at minus one diode drop, causing excessive power dissipation, at the least. Most op-amps won't be killed by this, at least at room temperature since modern op-amps typically have current-limiting but it may get quite hot. It will 'work' (in the sense that the output matches the diagram) but...
Just leave out D2 and it will be okay-ish. The recovery time of the op-amp from being saturated will impact how high a frequency can be accurately rectified. Simulation does not always accurately model that condition. A 10MHz op-amp may take tens of microseconds to come out of saturation.
Unfortunately, "example" circuits on datasheets run the gamut from circuits contrived to sell a particular proprietary product, to clever circuits that are excellent (anything from Pease or Williams R.I.P.\$^2\$, for example), to amateurish garbage.
Due to the negative feedback the op amp will always try to keep its inputs at the same voltage -- i.e., the input voltage, since the non-inverting input is connected directly to the input.
For positive input, the voltage on the right node of \$R_1\$ (the inverting input) is positive so a current flows through \$R_1\$ from right to left. Since no current flows into the op amp's input, the same current must flow from right to left through \$R_2\$ -- which means the output is at a higher voltage than the input (or equal to the input, if \$R_2 = 0\$). The op amp drives the output to the desired voltage by sourcing a current out of its output (the anode of \$D_1\$), and this current flows through \$D_1\$. The op amp's output will be a diode drop higher than the output voltage.
For negative input, the op amp tries to drive the voltage at its inverting input to the same negative input voltage. This would require a current flowing through \$R_1\$ from left to right, which in turn requires a current flowing through \$R_2\$ from left to right. However, \$D_1\$ blocks such a current in that direction. The op amp sinks current into its output in order to try to pull the output voltage down, but again the op amp's output current is blocked by \$D_1\$. Instead, the op amp's output current flows through \$D_2\$ from ground. This isn't really a short circuit because most op amps can only source or sink tens of milliamps at their output. The op amp's output voltage would thus be stuck at a diode drop below ground in this case.
