[ACPlayer]

I understand the point about how to account for odds when there are two cards to come. However, my point is that the backdoor draw is one out whether you are allin or not, on the flop

In your note you state:
"But if you could go all-in on the flop, then this is 6 outs with 2 cards to come, and you would only need pot odds of 1-(41/47)(40/46) = 4.1-1. "

This calculation is actually only true if the 6 outs were six outs on the turn and on the flop. The backdoor out is a pseudo out as it requires the first card to be one of the backdoor out cards. THat is if you are not all-in the number of outs on the turn could reduce to five if the allin outs dont hit.

So, because on the flop we are talking about the backdoor out being about 4% and knowing that one out twice is about 4% we are converting the backdoor chance into approx one out on the flop (this by itself assumes that 2 cards are to come). It does not matter whether you are allin or not. On the turn we either have 5 outs or 14 outs. Note that for accurate calcualation we would need to confirm that the backdoor outs are not already counted in the other outs (for example the 2 pair outs are not counted) but that is a secondary effect.