Here is a general answer that mostly just sums up the consequences of Schur–Zassenhaus.
Hypothesis and goal: Suppose G is a group with a normal Hall subgroup N, that is the order and index of N are coprime. We want to describe G as extension of a group of the same order as N, and determine all descriptions up to isomorphisms of G.
Observation 1: N is the unique subgroup of G of the same order as N. In particular, the isomorphism type of N is determined, and the specific copy of N in G is determined as a subset of G, and up to an element of Aut(N) for the embedding.
Observation 2: G is a semi-direct product of Q ≅ G / N, and Q is determined up to conjugacy in G (Schur–Zassenhaus, or Sylow's theorem in the 140 case). In particular, the isomorphism type of Q is determined, the specific copy of Q in G is determined up to conjugacy by an element of N, and and the embedding into a specific copy is determined up to an element of Aut(Q).
Observation 3: The action of Q on N is determined up to a pair of elements of Aut(Q)×Aut(N). In particular if N is abelian, then this is the same as an Aut(Q)-conjugacy class of Q-module structures on N.
Observation 4: The reverse is true: given an Aut(Q)-conjugacy class of Q-module structures on N we get an isomorphism class of G, the semi-direct product.
Result: Hence we have a one-to-one correspondence between isomorphism classes of groups G with normal abelian Hall subgroup N and quotient Q ≅ G / N and the Aut(Q)-conjugacy classes of Q-module structures on N.
This uses that N is a normal Hall subgroup on several occasions to radically simplify the arguments. At the very least, I always work with characteristic N, but in this case we have vanishing first and second (and third) cohomology which simplifies several steps.
More down to earth comments:
Hence we just count homomorphisms from Q to Aut(N) up to Aut(Q)×Aut(N) conjugacy. For G of order 140, N of order 35, that means N is cyclic of order 35, and Q is either cyclic or elementary abelian of order 4, and there are 6 and 5 such classes of homomorphisms, respectively. They are listed in the pdf you mentioned.
For Q cyclic, you can distinguish each of the actions by what it does on the Sylow 5-subgroup and the Sylow 7-subgroup as the effect can be described intrinsically as: centralize, invert, or (in the case of the Sylow 5-subgroup) "do an order 4 thing". The three possibilities on the (normal) Sylow 5-subgroup combine in all possible ways (due to CRT) with the two possibilities on the (normal) Sylow 7-subgroup, giving both at most and at least six distinct groups.
For Q elementary abelian, things are a little more complex, but basically on the Sylow 5-subgroup it must either centralize, or have two of its elements invert. If it centralizes, then we have the same two possibilities on the Sylow 7-subgroup ( so (1)(2) + (1)(?) groups so far). If it inverts, then the question is how many of those inverters of the 5 remain inverters of the 7. Answers are 0, 1, 2 giving a total of (1)(2) + (1)(3) = 5 groups.
One should get similar results for groups of order 4pq where p, q are primes with p not dividing q−1 and q not dividing p−1, with exactly one of p, q equivalent to 1 mod 4.
For instance, there are 11 groups of order (4)(5)(19) = 380 and of order (4)(7)(13) = 364. If both p and q are equivalent to 1 mod 4, then you get an extra possibility on the cyclic Q side too: (3)(3) actions individually, but now it matters if the order 4 on the 5 and the order 4 on the 7 match, giving (3)(3)+1 = 10 groups with cyclic Q and 5 with elementary abelian Q.
You're off to a good start. Here's how it goes.
$M$ is isomorphic to either $\mathbb Z_{21}$ or the Frobenius group, $F_{21}=\left<a,b\mid a^7=b^3=e,ba=a^2b\right>$. And clearly, $P_2\cong\mathbb Z_2$, say, $P_2=\left<c\right>$.
If $M\cong\mathbb Z_{21}$, say $M=\left<a\right>$, then $cac^{-1}=a^t$ for some $0\le t<21$ since $M$ is normal, thus $ca=a^tc$. Since $a=c^2a=ca^tc=a^{t^2}cc=a^{t^2}$, $t^2\equiv 1\pmod{21}$ so that $t$ is either $1$, $8$, $13$ or $20$. These four cases yield the groups $\mathbb Z_{42}$, $\mathbb Z_7\times S_3$, $D_7\times\mathbb Z_3$, and $D_{21}$ respectively.
Now suppose $M\cong F_{21}$. Then the map $\sigma:F_{21}\to F_{21}$ sending $x\to cxc^{-1}$ is an automorphism and $\sigma^2$ is the identity map $\iota$.
Now, $\sigma(a)=a^t$ for some $1\le t\le 6$ (because those are the only elements of $F_{21}$ of order $7$) and $\sigma^2=\iota$ implies $t^2\equiv 1\pmod 7$ so that $t$ is either $1$ or $6$. If $t=1$, then $\sigma(b)=a^ib$ for some $i$ (it can be shown that $\sigma(b)a=a^2\sigma(b)$ which would fail if $\sigma(b)=a^ib^2$); $\sigma^2=\iota$ implies $2i\equiv 0\pmod 7$, so that $\sigma(b)=b$ and $\sigma=\iota$. In this case, $M$ and $P_2$ commute, and $G\cong F_{21}\times\mathbb Z_2$.
Now suppose $t=6$. Then it doesn't matter what $\sigma(b)$ is; $\sigma^2=\iota$ would follow anyway, because if $\sigma(b)=a^ib$, then $\sigma(a^ib)=\sigma(a)^i\sigma(b)=(a^6)^ia^ib=a^{7i}b=b$. Hence, $\sigma^2(a)=a$ and $\sigma^2(b)=b$, so that $\sigma^2=\iota$. It can be shown that there are $7$ possibilities for $\sigma$ and each yields an isomorphic result for $G$, so suppose $\sigma(b)=b$. Then $bc=cb$, $\left<bc\right>$ is a cyclic subgroup of order $6$, $bca=a^5bc$ and $G$ is isomorphic to the holomorph $\mathbb Z_7\rtimes\mathbb Z_6$.
Summarizing, the six groups of order $42$ are $\mathbb Z_{42}$, $\mathbb Z_7\times S_3$, $D_7\times\mathbb Z_3$, $D_{21}$, $F_{21}\times\mathbb Z_2$ and the holomorph $\mathbb Z_7\rtimes\mathbb Z_6$. It can be shown quite feasibly that no two of those are isomorphic.
Best Answer
Here are some hints to help along:
Thus there are 4+6=10 possible groups. It is fairly easy to show all are distinct.