In this chapter, page 59, we can read
In problems 1-8, express $\Delta y$ and $dy$ as functions of $x$ and $\Delta x$, and for $\Delta x$ infinitesimal find an infinitesimal $\epsilon$ such that $\Delta y=dy+\epsilon\,\Delta x. \\\\ \;\;\;1\;\;\;\;\;\;\;\;y=x^2$
Correct answer: $\Delta y=2x\Delta x+\Delta x^2$, $dy=2x\Delta x$, $\Delta y=dy+\Delta x^2$.
I understand that $$y=x^2$$ $$y+\Delta y=\left(x+\Delta x\right)^2$$ $$\Delta y=2x\Delta x+\Delta x^2,$$ $\Delta y=2x\Delta x+\Delta x^2.$
But where do $dy=2x\Delta x$ and $\Delta y=dy+\Delta x^2$ come from? How do I derive them mathematically?
That is true by definition. In your text, the author defines that $$... dy = f'(x)\triangle x$$,
at the end of page 55.
EDIT: Well, recall the tangent line equation at the point $(a,b)$: $$T(x) = f'(a)(x-a)+f(a).$$
Then, $dy = T(x+\triangle x) - T(x) = f'(x)\triangle x$, because $dy$ is the change in $y$ along the tangent line at the point which you are starting that small increment $\triangle x$. There is a very well picture illustration in your text, on page 56. Once you have this, $$\triangle y = 2x\triangle x +x^2 = f'(x)\triangle x+x^2 = dy+x^2,$$ if you are working on the function $f(x) = x^2.$