Well assuming that we go for the approach where we form build a XXcross (or pseudo XXcross) after the second 4 edges have been built:
The main idea is to 1) Build 2 edges as fast as possible and 2) take care of the remaining two edges in one alg (after necessary setups)
Here are some options for 1)
a) (This is the most popular choice for me) Try to scan for tredges which contain a good midge+wing pair. Move this to the FL or FR slot and perform a slice-replace-slice back wing cycle so that at least one edge has been built + another good midge+wing pair has been made. Repeat this for the second edge if necessary.
b) Or try to look for a good pair of wings on a tredge. Perform a similar procedure like with 1a except you cycle midges around for your first edge. Don't forget that you can use this cycle to help: F' M' U2 M F or the mirror: F M' U2 M F'.
c) If neither of the above starts are available try to look for a checkered midge+wing pair (a pair with wing next to a midge which needs flipping). Try to find the remaining component and perform a slice-flip tredge-slice back wing cycle to build the edge. Another thing you can do if you spot two checkered midge+wing pairs is: Place them opposite to each other and perform a slice-flip tredge slice back
midge cycle to transform them into two good midge+wing pairs.
d) If all of the above fails well I guess you're left with the worse case possible. So you have no good pairs or checkered pairs anywhere. I'm not sure if I have a decent approach for these cases. Normally I try to form two good midge+wing pairs, (sometimes wing-wing), and finish off with approach a) (sometimes b)
This is how I generally take care of 2)
if I have parity (the last 2 edges):
a) If both edges are in FL and FR, simply do F' x (parity alg) x' F. Or R U R' F (parity alg) F'(It depends on the case obviously). Of course, the mirror setups are also options.
b) If there is one in FL or FR and one on the top layer:
i) FL: Move the one on top to UB and do F (alg) F' OR move the one on top to UR and do L' U'/U (alg) L
ii) FR: Move the one on top to UB and do F' (alg) F OR move the one on top to UL and do R U'/U (alg) R'
c) If both are on top but adjacent to each other:
i) Do a simple 4 move setup like R U R' U' or R U' R' U etc. and execute the algorithm
ii) Move both of them to UL and UF and do L' B' (alg) B L OR F R U'/U R' (alg) F'
iii) Move both of them to UR and UF and do R B (alg) B' R' OR F' L' U'/U L (alg) F
d) This isn't really another case but if I am left with OLL parity in FL or FR, I usually just continue my 333 solve until it appears on the top layer and I tend to take care of it after completing F2L.
As you can see in the 2nd and 4th examples, I do not solve 2 F2L pairs because I spotted an easy cycle for next edge so I decide to build the last 4 edges without the solving 2 F2L pairs. This isn't my usual approach to building the 4 edges, but I think it's worth it for easy cases. However I still take (almost) the same approach for building the last 2 edges.
I realise some of this is obvious, but I hope this helps you and others out
