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.
Best Answer
Edit: There is a big leap of logic in this answer of mine - the assumption that all elements of order $5$ are a power of $b.$ This is true, but definitely requires proof. I'd honestly delete this answer if I could, but it's status as selected answer makes that impossible.
Consider it this way. Let $a$ be an element of order $2$ and $b$ an element of order $5$. Show that $a^ib^j$ is all the elements of the group, with $i=0,1$ and $j=0,1,2,3,4$. Then $aba^{-1}$ is of order $5$, so it is equal to $b^j$ for some $j$. But then $b=ab^{j}a^{-1}=(aba^{-1})^j=b^{j^2}$. So $j^2\equiv 1\pmod 5$. What are possible values for $j$?