What does it mean for order of an element to be finite? What's finite number? 
Let $D$ be the set of all elements of finite order in an Abelian group $G$. Prove $D$ is a subgroup of $G$.
Proof: Let $a, b \in D$ such that $\operatorname{Ord}(a) = n$ and $\operatorname{Ord}(b) = m$. Then $\operatorname{Ord}(a^{-1}) = n$. Since $G$ is Abelian, $(a^{-1}b)^{mn} = (a^{-1})^{mn}b^{mn} = e$. Thus $\operatorname{Ord}(a^{-1}b)$ is a finite number so that $a^{-1}b \in D$. Thus $D$ is a subgroup $G$.

What's a finite number? What does it mean in this context?
 A: The order of an element is the power $p \in \Bbb{N}$ such that $a^p=1$. However, sometimes, there is no power such that $a^p=1$. For example, take the group $\Bbb{Q}^*$ under multiplication. No matter how many times you multiply $2$ by itself, there is no way you are going to get $1$. Therefore, $2$ is said to have an infinite order under $\Bbb{Q}$.
Thus, if an element $a$ has a finite order, there exists some $p \in \Bbb{N}$ such that $a^p=1$. However, if an element $a$ has an infinite order, there is no way to get to $1$ by taking $a$ to some power.
A: General theory
Let $g\in G$, where $G$ is a group. Then the set
$$
\langle g\rangle=\{g^n:n\in\mathbb{Z}\}
$$
is the smallest subgroup of $G$ with $g$ as a member. We can also see $\langle g\rangle$ as the image of the group homomorphism
$$
\varphi_g\colon\mathbb{Z}\to G,\qquad \varphi_g(n)=g^n
$$
By the very definition, $\langle g\rangle$ is a cyclic group and by the homomorphism theorem,
$$
\langle g\rangle\cong \mathbb{Z}/\ker\varphi_g
$$
Since $\ker\varphi_g$ is a subgroup of $\mathbb{Z}$, there is a unique $k\ge0$ such that $\ker\varphi_g=k\mathbb{Z}$.
If $k=0$, then $\langle g\rangle\cong\mathbb{Z}$ is infinite and $\varphi_g$ is injective. Therefore the only $n\in\mathbb{Z}$ such that $g^n=1$ is $n=0$. In this case $g$ is said to have infinite order.
If $k>0$, then $|\langle g\rangle|=|\mathbb{Z}/k\mathbb{Z}|=k$ and $k\in\ker\varphi_g$, so $g^k=1$ and also $g^m\ne1$ for every $m$ with $0<m<k$. Thus $k$ is the least positive integer such that the relation $g^k=1$.
In the case $k>0$, $k=|\langle g\rangle|$ is called the order of $G$ and $g$ is said to have finite order.
By the argument above, elements $g$ of finite order are characterized by the existence of a positive integer $m$ such that $g^m=1$, the least of which is the order.
Your theorem
The proof of you theorem can be simplified by not mentioning the order at all.
Let $a,b\in G$ have finite order. Then there exist $m>0$ and $n>0$ such that $a^m=1$ and $b^n=1$. If we set $t=mn$, we have $t>0$ and $a^t=1=b^t$.
Then $(ab^{-1})^t=a^t(b^{-1})^t=1\cdot(b^t)^{-1}=1$, so $ab^{-1}\in D$.
The first equality above exploits $G$ being abelian. Since clearly $1\in D$, we have proved that $D$ is a subgroup.
Comments
The textbook should have said “$ab^{-1}$ has finite order”, rather than “the order of $ab^{-1}$ is a finite number”. Such slips are common, however.
Your textbook might have defined the order of $g\in G$ as the least positive integer $k$ (if it exists) such that $g^k=1$. The definition is of course equivalent to the above one, but I find that using the homomorphism theorem is clearer.
