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.
Note that $20 = 2^2 \cdot 5$.
By Fundamental theorem of Finitely Generated Abelian Group, there are two distinct abelian groups of order $20$: $\mathbb{Z}_{20}$ and $\mathbb{Z}_{10} \times \mathbb{Z}_2$.
Now let $G$ be a nonabelian group of order $20$. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow $5$-subgroup $H \cong \mathbb{Z}_5$. Now let $K$ be any Sylow $2$-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order $20$ is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.
Let $K = \mathbb{Z}_4 = \langle x \rangle$. There are four distinct homomorphisms $K \rightarrow \mathsf{Aut}(H)$.
If $\varphi_1(x) = 1$, then $\varphi_1$ is trivial; this contradicts the nonabelianicity of $G$.
If $\varphi_2(x) = \alpha$, then $\mathbb{Z}_5 \rtimes_{\varphi_2} \mathbb{Z}_4 $is indeed a nonabelian group of order $20$.
If $\varphi_3(x) = \alpha^2$, then $\mathbb{Z}_5 \rtimes_{\varphi_3} \mathbb{Z}_4$ is indeed a nonabelian group of order $20$. Moreover, since $\mathsf{ker}\ \varphi_3 \cong \mathbb{Z}_2$ and $\mathsf{ker}\ \varphi_2 \cong 1$,$ H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K$.
If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $\mathbb{Z}_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$.
Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup.
Suppose now that $K = \mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : \mathbb{Z}_2^2 \rightarrow \mathbb{Z}_4$ is determined uniquely by $\psi(a)$ and $\psi(b)$, and is indeed a homomorphism provided $|\psi(a)|$ and $|\psi(b)|$ divide $2$. We thus have $\psi(a)$, $\psi(b) \in \{ 1, \alpha^2 \}$, for a total of four choices.
If $\psi_1(a) = \psi_1(b) = 1$, then $\psi_1 = 1$, contradicting the nonabelianicity of $G$.
If $\psi_2(a) = \alpha^2$ and $\psi_2(b) = 1$, then $\mathbb{Z}_5 \rtimes_{\psi_2} \mathbb{Z}_2^2$ is indeed a nonabelian group of order $20$.
If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^2$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$.
If $\psi_4(a) = \alpha^2$ and $\psi_4(b) = \alpha^2$, then $\psi_4 = \psi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$.
Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup.
In summary, the distinct groups of order $20$ are as follows. We let $\mathbb{Z}_5 = \langle y \rangle$, $\mathbb{Z}_4 = \langle x \rangle$, and $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.
$Z_{20}$,
$Z_{10} \times Z_2$,
$Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.
$Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$
$Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.
(Source: Crazyproject)
Best Answer
The systematic approach is to write down the factorisation $ 900 = 2^23^25^2 $. Then, for each prime factor, find all partitions of its exponent. Combine all these partitions to get the following list (I write $C_n$ instead of $\mathbb{Z}_n$ because I'm lazy):
$\begin{array}c C_4 \times C_9 \times C_{25} ( \cong C_{900} ) ,\\ C_2 \times C_2 \times C_9 \times C_{25} ,\\ C_4 \times C_3 \times C_3 \times C_{25} ,\\ C_4 \times C_9 \times C_{5} \times C_5 ,\\ C_2 \times C_2 \times C_3 \times C_3 \times C_{25} ,\\ C_2 \times C_2 \times C_9 \times C_{5} \times C_5 ,\\ C_4 \times C_3 \times C_3 \times C_{5} \times C_5 ,\\ C_2 \times C_2 \times C_3 \times C_3 \times C_{5} \times C_5. \end{array}$
The structure theorem for finitely generated abelian groups, also known as fundamental theorem, confirms that these are indeed all abelian groups of order $900$ and that they are pairwise non-isomorphic.