For some background information, Eulerian here refers to starting and ending on the same vertex (i.e. being closed).
Also, the corollary below uses Theorem 3.4 which states a graph is Eulerian iff each edge lies on an odd number of cycles.
$\textbf{Question:}$ What is the reason that $s(G)\equiv \displaystyle\sum_{i=1}^rs_i$ holds true at the very bottom of the proof below?
This corollary is found on page 68 here.
Note I do understand the following in the proof:
$\text{ Set of All Cycle Decompositions of G}=\{\text{Every Cycle Decomposition of } G_i \cup C_i\}_{i=1, 2, ..., r}$
and that $s_i=|\text{Every Cycle Decomposition of } G_i \cup C_i |$ in consideration as to what is inside the set above.
At a glance, $s(G)=\displaystyle \sum_{i=1}^r s_i$ holds true by simply adding up each cycle decomposition at a time to get the total number.
However, wouldn't there be a chance that a cycle decomposition of $G_i$ $\cup C_i$ might be equal to a cycle decomposition of $G_j\cup C_j$ where $i\neq j$ which seems like a problem.
The problem that concerns you would arise only if $C_i$ appeared in a cycle decomposition of $G_j$ for some $j\ne i$. This cannot happen, however, because $C_i$ and $C_j$ both contain the edge $e$ chosen at the start of the proof: no cycle in a cycle decomposition of $C_j$ can contain $e$, since it belongs to $C_j$, and therefore no cycle in a cycle decomposition of $G_j$ can be $C_i$, which also contains $e$.