I did a LT Spice simulation but need help understanding why the diode is conducting on part of the positive cycle. Conducting on the negative side of the cycle makes sense to me. The input waveform is in green and the rectified voltage is shown in red. Spice sim is also attached to this post.

My understanding is the diode only conducts when: V(A)-V(B) ≥ Vf

For example: The Vf of the diode is .5V. At the start of the positive signal cycle lets say the voltage just reaches .4V. As I understand it, the diode still isn't conducting so the portion labled "rectified should still be 0 right? Following V(A)-V(B) ≥ Vf that would mean 0 - .4 ≥ .5V which would mean the diode shouldn't be conducting. Clearly my understanding is off but I can’t figure out why. If this was the result of a reverse voltage spec it would be a .5V constant for all of the positive cycle so I know that's not it and would apprecaite additional insight.

 

You are assuming an ideal diode behavior which real diodes are not.
One of the important things to learn when designing circuits is to be aware of the non-ideality of all real components, which is not always emphasized in school.

The reverse current is likely due to the junction capacitance of the diode (below for the 1N5818):
You can reduce that effect by using a smaller load resistance (below) or a lower-current (smaller) Schottky diode, which would have a lower capacitance.
10kΩ gives a very small load current for a 1A rated diode, so the small current from the capacitive feedthrough is readily seen.