Many of the E2L pair algorithms are not ideal in a speedsolve due to ergonomics, unless they result in triplet solves where the benefit of a multi-edge solve is so huge that it is okay to suffer a delay from setup and a less than perfectly ergonomic sequence. The key to E2L is to learn intuitive pair solving, sometimes called cycling.
Here are some examples. Hold the cube with orange on the front and white on the top:
Scramble 1: L U M U' L' M U M' U'
We want to solve the green white edge (at DF) into the UR slot. We see that the blue-red edge is in the UL slot. We want to solve both the green-white edge and the blue-red edge at the same time intuitively.
Step 1: Prime the blue red edge with U M U'. Now the UR slot has been 'primed' and holds the blue-red edge, ready for an easy pair. The green-white edge has been displaced to the BD position.
Step 2: Bring the green-white edge back to the DF position with an M' move.
Step 3: We can now solve the green-white edge and the blue-red edge at the same time. Do an L move to put the blue-red slot in the UL position, then solve the pair with U M' U'.
The above will work any time the UL slot contains a blue edge piece on the blue layer that is oriented; using this intuitive method you have just covered 3 possible E2L situations and this intuitive pair method is extremely fast and ergonomic (no F or B or D moves, just L/R/U/M).
Scramble 2: Hold the cube with orange on front and white on top, U M U' l U M U' M'
We see the blue-red edge is in the UR slot in an ideal position. We also see the blue white edge at the UF position. We can cycle-solve both the blue-white and blue-red edges at the same time. First do an M move to put the blue white edge at the DF position. Now solve the blue-red into UL with U M' U', and this simultaneously places the blue-white edge into the UR slot. Now we solve that one next; L' U M' U
Scramble 3: (white on top, orange on front) U' M U R2 U' M' U
We want to solve the green-yellow edge (at BD) into the UR slot. We see there is already a green-white edge there. We can cycle solve them both with no algorithm. First do U' M U, and this solves the UR slot with green-yellow; it also primes the UL slot with the green-white edge. Now quickly do an r2, then U' M' U to solve the green-white edge piece (I deliberately didn't solve the blue-white edge at the same time, this was possible but unlikely to occur so nicely in a real solve)
Scramble 4: (white on top, orange on front) U2 R' U' M' U R U2 R'
The blue face is solved. We want to solve the green-red edge piece into the UR slot. Green-red is at BD. However there is a green-white edge piece already in the UR slot and we can solve it intuitively. This situation is defined as trying to solve an edge piece where there is another edge piece of the same side already in the slot but disoriented. In this case it could be green-white (as it is), or green-orange too for example. It is all the same.
Solution: We set this up as a basic U/M/U style pair and with all intuitive solving it is about 'priming' edge pieces for an easy pair. First do a U2. Now you have sort of inverted the blue/green faces. The green-white edge piece is now at UL and in a very favorable primed position for a pair. The green-red edge piece is still at BD and now it needs to go into UL. The green-white edge piece will be displaced to UR, so we need to put the green white slot at UR with an R move. Now solve the pair with U M U'. The edges are now both solved and we need to 'restore' the cube. We restore it with R' U2. Once you understand this 'style' of flipping the top slice with U2 to 'prime' a pair that exists on the same side, you can use it to solve any similar situation.
Scramble 5: (white on top, orange on front) F U' M2 U F' M'
We want to solve the blue-white edge (at DF) into UL. We see that UR (green-orange) and FR (green-white) are both disoriented and swapped. In a real situation this might happen, but more than often only one of UR and FR will hold such an edge piece. This is more of an algorithm but it is quite intuitive. We do M F U' M2 U F' and it solves all three pieces. This can be used in many situations such as this:
(white on top, orange on front) F U' M2 U F' U' M' U R
We want to solve the blue white edge at DF, into UL. We seed the green-white edge at UR would need to go to BR.
First do an R'. Now we are in our familiar situation. By doing our mini-algorithm we know the UR and FR will be swapped and oriented. We see green-white at FR, and this will swap into UR and its orientation will flip. So we execute our mini-algorithm M F U' M2 U F' and both edges are solved. This works any time either of UR/FR need to be swapped.