Remember the definition of what it means for an integer to divide another integer.
The following are equivalent
$a\mid b~~~$ (read aloud as "$a$ divides $b$")
There exists some integer $k$ such that $b=a\cdot k$
$b\equiv 0\pmod a$
$b$ is an integer multiple of $a$
$a$ is a divisor of $b$
among others...
To prove the claim in the beginning of your post written above ($a\mid b\implies Div(a)\subseteq Div(b)$), we will use the claim in the bottom $(a\mid b$ and $b\mid c$ implies $a\mid c$)
The proof for the second claim, I will rewrite here with hopefully enough words that you can understand it.
Claim: $a\mid b$ and $b\mid c$ implies $a\mid c$
Proof:
Since we wish to prove that knowing $a\mid b$ and $b\mid c$ would be enough to conclude that $a\mid c$, we begin by assuming that our hypothesis holds true. I.e. we suppose that $a\mid b$ and we suppose that $b\mid c$ are both true statements.
By the definition of "divides" this means that since we assumed $a\mid b$ to be true, we can find some integer, I'll call it $k$, such that $b=a\cdot k$.
Also, again by the same definition, since we assumed $b\mid c$ to be true, we can find some integer, I'll call this one $n$, such that $c=b\cdot n$.
(Note: there is no significance to what I name these integers that I am saying exist. I just need to remember what role each of these named integers play in my proof. To keep them separate, I needed to name them different things.)
So, so far we know that $b=a\cdot k$ and that $c=b\cdot n$ for some integers $k$ and $n$.
We notice now that since $b=a\cdot k$, I can use this expression for $b$ when writing the expression for $c$. We have then $c=b\cdot n = (a\cdot k)\cdot n$ as a result of making the replacement. Now, by the associative property of multiplication we have $c=a\cdot (k\cdot n)$
Finally, noting that since $k$ and $n$ are both integers, so too is $(k\cdot n)$ since multiplication of two integers produces another integer. We have then successfully written $c$ as an integer multiple of $a$, implying by definition that $a\mid c$.
This is often written much more concisely with fewer words. E.g. Let $a\mid b$ and $b\mid c$. Then there exist integers $k,n$ such that $b=ak$ and $c=bn$. Then $c=a(kn)$ and thus $a\mid c$.
$~$
As for where they got $a\cdot (n_1\cdot n_2)$ from in the proof in your post, they simply knew ahead of time that it would work. They wrote their final line backwards compared to how I wrote it in my long proof above, but that doesn't make it any less correct.
Using this for your original problem:
Claim: $a\mid b$ implies $Div(a)\subseteq Div(b)$
Proof:
To prove one thing is a subset of another, we show that any element in the set on the left must also be an element on the right.
Let $k\in Div(a)$. Then $k\mid a$. Since $k\mid a$ and $a\mid b$ by the previous proof this implies $k\mid b$ and therefore $k\in Div(b)$.
This proves then $Div(a)\subseteq Div(b)$.
