I think the combinatorial argument could be more useful if made more clear.  I want to make sure I am using it correctly; consider this answer an extension of the question and asking for other uses of this argument in the literature.

What I call $u(m)$ (and the authors call $u(m,X)$) is a nice counting function.  For sufficiently large positive integers $m \leq X$, $u(m)$ takes on the values $0,1,$ and $2$.  One chooses the exponent $1/e$ to get the summatory function $\sum_{1\leq m \leq X}  u(m)$ up to a value close to and above $X$, and indeed $c_8$ does not need to be much smaller than $1/e$ to accomplish this, so one can make the argument with an explicit choice. One can also define $L$ so that the sum is within a small constant times $X^{1/e}$ from $X$. (We can also vary $X$; this will be crucial later.)

For arbitrary $ X$, one can show the existence of the stated $m$ as follows.  Consider for any positive integer $j$ less than $X$ the largest $s$ such that $\sum_{1 \leq i \leq r} u(j+i) \geq r$ holds for $1 \leq r \leq s$ but does not hold for $r=s+1$.  Indeed, for small $j$ outside of $L$ $u(j+i)$ is $0$ for small $i$, and so for these $j$ $s$ is $0$. One can partition $[1,X]$ into disjoint intervals of the form $[j,j+s]$, and there will be at least $c_8X^{1/e}$ many of these intervals, which means (since the summatory exceeds $X$) that the excess $u$ values pile up later on in the interval. $m$ is the largest of these $j$'s and has to handle all the excess "missed out" by the smaller $j$'s, so we must have $m \in (c_8X^{1/e}, X - c_8X^{1/e})$. This is argument has less control if you replace $1/e$ by something smaller, say $1/10$.  You can show the existence of $m$, but $u$ would become more bumpy and you might not prove as large a value of $s$ as here ($m$ might end up too close to $X$.)

As currently written, it is not clear if the authors expect $\sum_{1 \leq i \leq X+1-m} u(m+i) \geq X +1 -m$ to fail. Indeed, if it does not fail, then $v(m,X-m+1)$ may be larger than $ X-m+1 $ just because $W(m,X-m+1)$  might have that many prime factors bigger than $c_8X^{1/e}$ alone, and still more small prime factors.  Then $f_0(m)$ would be larger than $X-m+1$, and $m$ might not be one of the infinitely many $n$ highlighted in Theorem 2.  More importantly, the suggested value of $n$ of $X-t$ or $n= X - c_6X^{1/e}$ might also not satisfy $f_0(n) \lt c_6X^{1/e}$ because we found a large interval full of primes and semiprimes containing $X$. This issue as well as having to fill in details like "where is $n$?" prompts the question.

If we can guarantee that the last inequality fails for $t= X-m+1$, then I can finish the proof easily.  Choose $c_6$ and $t$ as suggested, and choose $n=X-t$. Then $W(n,t)$ has fewer than $t$ factors coming from $L$ by the construction of $m$ and the guarantee. If also $c_6$ is larger than $c_8$ and $n \gt m$, then some of the factors from $L$ are repeated (so not distinct) and since we can vary $t$ between $1$ and $X-m$, we can choose $c_6$ so that there are more repeats than primes below $c_8X^{1/e}$, and now we get finally an $n$ that gets close to the goal, having $v(n, X-n) $ less than $c_6X^{1/e}$. To get the goal requires a little more tweaking of $c_6$, but we have made assumptions on $X$ and $m$, so let's address those.

So let us not choose arbitrary $X$.  Let us choose $X' \geq X$ so that (now renaming $X'$ to $X$) we can assert the existence of $m \lt X$ above where the inequality in $t$ above not only works for $1 \leq t \leq X-m$ **and also** fails at $t=X+1-m$.  This is essentially the partition argument above, except we choose $X'$ to be $m+s$ (or $m+s+1$, whichever works) and rename it to $X$. We can also tweak $c_8$ and $L$ correspondingly to get this to work.

Here is the key point: for all $n$ in $(m, X)$ now we have $\sum_{1 \leq i \leq X-n} u(n+i) \lt X-n$, and now we have the inequality pointing in the right direction (because of the failure for $m$ at $t=X+1-m$).  Now we can have our pick of $n$ to get the small primes in $W(n,X-n)$ outnumbered by the repeats in $L$. Even if $m$ is too small, we can start "backing up" (by choosing $m' \lt m$ and renaming) and find a good value for $n$ less than the original $m$. There are more details to consider, but now I see a light at the end of the tunnel, and it does not look like an oncoming train.

I can understand leaving detail to the reader, but I think the reader deserves the bone of the phrase ("and let $n=X-t$").  Unless, of course, I missed something simpler.

Gerhard "Maybe Can Hitch A Ride?" Paseman, 2017.09.08.