1st Block: F' B2 D2 U2 y'...
Roux is preferably rotationless.
1st Block: F' B2 D2 U2 y' B R' F2 R U2 L' U L
Whoa F2L.
U' L is enough
2nd Block y R' U R2 U R' M U M' y L F2 L'...
I'd do R U R2 U'.
No rotations and all that.
Easily continues into (R' U R) (r2' U' r)
General rule for 2nd block is to use only RrMU moves, with no rotations.
2nd Block y R' U R2 U R' M U M' y L F2 L' y' M U2 y R U' R' U2 R U R' y'
Whoa F2L again.
Had to rotate for it too.
Try (r' U r) U2 (r' U' r)
L6E: U M' U' M U F2 M F2 M'
Wow.
Why not just M' U2 M U2?
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L6E: U M' U' M' U2 U F2 M F2 M' ( I found this after 5 min. It's not my quick lookahead)
Why?
It's almost what I got.
Here was my thought-process.
4a should be simple Moo-case.
I was tracking UL and UR edge for 4b (I usually start looking for them at CMLL), so I knew to do the Moo with a U' to align the edges so they go on the same face at the end of the moo.
Plus, the edge was between the "side corners" (same or opposite color as the edge, L/R color), so it's one of those familiar combined 4a+4b steps that you should always look out for to avoid wasted moves in L6E (just means I can insert directly with M' at the end of moo, instead of bringing both edges to the D face with an M).
For 4c, I saw the BU sticker during 4b, matched the FU sticker, so I knew that if it was a Fuu, I had to do an M or M' first (opposite stickers mean either U2 or M2 first).
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Also, not sure if it was just a coincidence, but no need to solve the pairs around the LD/RD edge (ala F2L).
Building the same first block:
I see I can either connect
1) ULB corner with UR edge (after flipping the edge)
2) Connect BRD corner with FU edge, with B R2 (and the R2 also flips the edge for #1)
And so you get B R2 B F' D2 R U' B
Probably not optimal, but I just wanted to show a pair that includes the LD edge instead.