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If group $G$ is isomorphic to $\mathbb Z_{k_1}\oplus\dots\oplus\mathbb Z_{k_r},$ such that $k_i\mid k_j$ and $i

Or am I wrong ?

Or if I'm wrong, what are the possibilities? Can you give a concrete example?

In text below, why does the author search for the order (or cardinality) of the group, it is already isomorphic to $\mathbb Z_3\oplus\mathbb Z_3$ ? (book: Elliptic Curves, Number Theory and Cryptography, Lawrence Washington)

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    That's correct. – 2017-01-05
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    The order of $\mathbb{Z}/n\oplus \mathbb{Z}/m$ is always $nm$, independent of the possibility that $n\mid m$ or not. – 2017-01-05
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    If $A$ and $B$ are finite groups, then $|A\times B|=|A|\,|B|$; the notation $\oplus$ is just the same as $\times$, used for abelian groups. – 2017-01-05
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    "Why does the author say that?" Perhaps the author is not assuming that the reader really knows any abstract algebra. Or perhaps the author wanted to illustrate the poverty of the method being discussed, and then contrast it to the kind of thing you can do with the *next* theorem, or some facts from basic algebra. It's hard to tell without further context. – 2017-01-05

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You are correct. And if you name the elements of $\Bbb Z_k$ as $0, 1, 2, \ldots, k-1$, then the elements of your group can be enumerated as $$ (u_1, \ldots, u_r) $$ where $$ 0 \le u_i < k_i $$ which shows that there are $\Pi_{i=1}^r k_i$ of them.