When the authors write "the proof for case 2 of Lemma 4.3 will go through", they mean that statement $g(n) = O\left(n^{\log_b a} \lg n\right)$ can be proven from $f(n_j) = O((n/b^j)^{\log_ba})$ using a proof that is very similar to that for case 2 of Lemma 4.3.
Basically, that proof is the following:
Substituting $f(n_j) = O((n/b^j)^{\log_ba})$ into equation (4.14) yields
$$
g(n) = O\left(\sum^{\left\lfloor\log_b n\right\rfloor - 1}_{j = 0} a^j \left(\frac n {b^j}\right)^{\log_b a}\right) \;\;\;\;\; (*)
$$ We bound the summation within the $O$ as in case 1, but this time we do
not obtain a geometric series. Instead, we discover that every term of the
summation is the same:
\begin{align}
\sum^{\left\lfloor\log_b n\right\rfloor - 1}_{j = 0} a^j \left(\frac n {b^j}\right)^{\log_b a}
& = n^{\log_b a} \sum^{\left\lfloor\log_b n\right\rfloor - 1}_{j = 0} \left(\frac a {b^{\log_b a}}\right)^j \\
& = n^{\log_b a} \sum^{\left\lfloor\log_b n\right\rfloor - 1}_{j = 0} 1 \\
& = n^{\log_b a} \left\lfloor\log_b n\right\rfloor \\
\end{align}
Substituting this expression for the summation in equation $(*)$ yields
\begin{align}
g(n)
& = O\left(n^{\log_b a} \left\lfloor\log_b n\right\rfloor\right) \\
& = O\left(n^{\log_b a} \lg n\right) , \\
\end{align}
Your solution is equivalent to
$$\sum_{i,j,k=1}^2g^i_{jk} $$
and no summation convention (I think it is a way to say "Einstein convention on repeated indices") appears, as pointed out by @Raskolnikov.
The aim of the exercise is to arrive at an expression with "repeated indices", i.e. an expression in which you use the summation convention. To do so, one needs to have no free index (like yours $i$, $j$ and $k$) and to contract-or produce pairs of- all indices.
It is clear that the starting expression has 3 indices: so 3 summations, or contractions are needed. The textbook begins to produce a summation considering at first the above index, called $i$. This is done introducing the vector
$$c=(c_1,c_2)=(1,1)$$
and realizing the sum $\sum_{i=1}^2 g^i_{j,k}$ (which, once again, uses no summation convention) as
$$g^1_{j,k}+g^2_{j,k}=\sum_{i=1}^2 g^i_{j,k}=g^i_{j,k}c_i,$$
for any $j,k$. On the rightmost r.h.s. of the above expression we use the Einstein convention, summing over $i$, the only repeated index. Please note that the length of $c$ is equal to $2$, i.e. the cardinality of the set $\{1,2\}$ of all the possible values for $i$.
We are left with 2 indices, so 2 more summations are needed. We repeat the above lines to produce the summation w.r.t. to $j$ through
$$g^i_{1,k}c_i+g^i_{2,k}c_i=\sum_{j=1}^2 g^i_{j,k}c_i=g^i_{j,k}c_ic_j,$$
for any $k$. Repeating the same trick with $k$, the only free index remaining, we arrive at
$$\sum_{k=1}^2g^i_{j,k}c_ic_j=g^i_{j,k}c_ic_jc_k,$$
and
$$\sum_{i,j,k=1}^2g^i_{jk}=g^i_{j,k}c_ic_jc_k.~~~(1) $$
On the r.h.s. of $(1)$ we have all repeated indices and we are done.
Best Answer