If $G$ is a solvable group, then every homomorphic image of $G$ is solvable (Fundamentals of Abstract Algebra by D.S. Malik, John M. Mordeson, M. K. Sen, page 229, 230).
Let $f$ be a homomorphism of $G$ onto a group $\bar G $.
Set $\bar H_i = f(H_i), i = 0, 1, ... , n$.
It is known $f$ is an epimorphism, so $f(H_{i+1})$ is a normal subgroup off $f(H_i)$
Also, $H_i \supseteq H_{i+1}$ implies that $f (H_i) \supseteq f (H_{i+1})$. Hence, $$\bar G =\bar H_0 \supseteq\bar H_1 \cdots \bar H_n=\{e\} \cdots (8.15)$$ is a subnormal series of $\bar G$. We now show that $f(H_i)/ f(H_{i+1}) = \bar H_i/\bar H_{i+1}$ is commutative.
Define $g : H_i \rightarrow \bar H_i/\bar H_{i+1}$ by $g(h_i) = f(h_i)\bar H_{i+1}$ · Since $f$ is an epimorphism, it follows that $g$ is an epimorphism of $H_i$ onto $\bar H_i/\bar H_{i+1}$·
Note that for any $$ h_{i+1} \in H_{i+1} \subseteq H_i, g(h_{i+1}) = f(h_{i+1})\bar H_{i+1} = f(h_{i+1})f(H_{i+1}) = f(H_{i+ 1})$$.
Hence, $H_{i+1}\subseteq \rm Ker \; g$. Thus, $g$ induces an epimorphism of $H_i/H_{i+1}$ onto $\bar H_i/\bar H_{i+1}$·
Since $H_i/H_{i+1}$ is commutative, it follows that $\bar H_i/\bar H_{i+1}$ is commutative.
Consequently, the subnormal series (8.15) is a solvable series, proving that $\bar G$ is a solvable group.
I have 3 issues in this proof, since it is a single proof and all 3 questions are related, I am compiling all 3 in this single post.
Question 1. Why $H_{i+1}\subseteq \rm Ker \; g$? Should not it be $H_{i+1}= \rm Ker \; g$?
Question 2. How does $g$ induce an epimorphism of $H_i/H_{i+1}$ onto $\bar H_i/\bar H_{i+1}$?
There is a natural homomorphism that maps $ H_i$ onto $ H_i/ H_{i+1}$, but how do(es) it or $g, f$ make $H_i/H_{i+1}$ onto $\bar H_i/\bar H_{i+1}$?
Question 3. How can we infer that $\bar H_i/\bar H_{i+1}$ is commutative from the fact previously found that $H_i/H_{i+1}$ is commutative?
NOTICE:
In the book there is a printing mistake, $ h_{i+1} \in \bar H_{i+1} \subseteq \bar H_i$ is written instead of $ h_{i+1} \in H_{i+1} \subseteq H_i$.
Q1: Equality is not true in general. The lesson here is probably that you should think of some simple concrete examples before moving on if things sound fishy to you.
Take $G = \mathbb{Z}$ and take $\bar{G} = \{0\}$. Take $f:G\to\bar{G}$ to be the $0$ map, i.e. $f(x) = 0$ for all $x\in\mathbb{Z}$.
Since $\mathbb{Z}$ is abelian the series with $H_0 = \mathbb{Z}$ and $H_1 = \{0\}$ will do. The corresponding images are $\bar{H_0} = \{0\}$ and $\bar{H_1} = \{0\}$.
So $H_0 / H_1$ is $\mathbb{Z}$ and $\bar{H_0}/\bar{H_1}$ is $\{0\}$. The map $g:H_0 \to \bar{H_0}/\bar{H_1}$ is the trivial map. That is $\ker(g) = H_0$.
Of course $H_1$ is a subgroup of $\ker(g)$, but it is obviously not equal to all of $H_0$.