The appearance of the term "Betti number" in your theorem should be seen as a definition. The Betti number of a finitely generated abelian group $G$ is the number of $\mathbb{Z}$ factors appearing in the decomposition claimed by your theorem, or differently stated, the rank of $G$ as a $\mathbb{Z}$-module.
You can also understand it as a sort of analogue to the dimension of a vector space. Indeed, tensoring $G$ with $\mathbb{Q}$ will convert it into a $\mathbb{Q}$-vector space whose dimension is exactly the rank/Betti number of $G$.
That is, for example, if $G=\mathbb{Z}^n$, then $n$ is the Betti number.
Another example is $G=\mathbb{Z}_{(p)}$. There is no $\mathbb{Z}$ factor, so the Betti number is $0$.
Now let us look at your two statements:
The first one is false: Take $G=\mathbb{Z}$ and $H=\mathbb{Z}\times \mathbb{Z}_{(2)}$. They both have Betti number $1$, but they are clearly not isomorphic.
The second one is true: if the Betti number was positive, then the group would have a subgroup isomorphic to $\mathbb{Z}$, in particular it would have infinitely many elements.
As noted by SpamIAm, the key to this will be the Smith normal form of a matrix. Basically, we write $N$ as the image of a matrix and then find an easier, similar matrix for which the solution is clearer.
Since we have 2 generators for $N$ we have a surjective homomorphism $\mathbb{Z}^2\to N$ given by the matrix $A$ of generators for $N$:
$$A=\begin{pmatrix}6&0\\0&0\\-3&8\\0&4\\3&2\end{pmatrix}$$
and $N=A\mathbb{Z}^2$. Now, this matrix its self is kind of messy, but it turns out that we can use a similar matrix and if we use a simple enough matrix the problem becomes easier.
Now you find the Smith normal form by using row and column operations to simplify as much as possible. I got the following matrix but you should check my work yourself:
$$A=\begin{pmatrix}6&0\\0&0\\-3&8\\0&4\\3&2\end{pmatrix}
\sim\begin{pmatrix}3&2\\0&0\\-3&8\\0&4\\6&0\end{pmatrix}
\sim\begin{pmatrix}1&2\\0&0\\-11&8\\-4&4\\6&0\end{pmatrix}
\sim\begin{pmatrix}1&0\\0&0\\0&30\\0&12\\0&-12\end{pmatrix}
\sim\begin{pmatrix}1&0\\0&6\\0&0\\0&0\\0&0\end{pmatrix}=A'$$
Now we have $\mathbb{Z}^5/N=\mathbb{Z}^5/(A\mathbb{Z}^2)\cong\mathbb{Z}^5/(A'\mathbb{Z}^2)$. But it should be easy to see that $A'\mathbb{Z}^2=\mathbb{Z}\times(6\mathbb{Z})$ which gives us
$$G\cong \dfrac{\mathbb{Z}^5}{\mathbb{Z}\times(6\mathbb{Z})}=\mathbb{Z}^3\times\mathbb{Z}_6$$
Best Answer
No, most non-finitely generated abelian groups are not products of cyclic groups. For instance, a product of infinitely many nontrivial cyclic groups is uncountable (since any product of infinitely many sets with more than one element is uncountable). So any countable abelian group that is not finitely generated cannot be a product of cyclic groups. A simple example of such a group is $\mathbb{Q}$.
The proof that $\mathbb{Q}$ is injective in those notes is totally wrong (it's not even correct if you assume $P$ and $Q$ are finitely generated; it seems to be implicitly assuming $P$ is a direct summand of $Q$). A correct proof involves extending $f$ from $P$ to $Q$ one element at a time by transfinite induction, using the divisibility of $\mathbb{Q}$.