As noticed in the comments, your first definition doesn't make sense. But don't worry, the fact that you thought that definition made sense is symptomatic of the way most people think about $dy$, as being an "infinitesimal diference in $y$".
The rigorous definition of $dy$ goes like this. First, let $y: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. Then, by definition, at $x \in \mathbb{R}$, the following limit exists and is called the derivative at x: $$\frac{dy}{dx}=\lim_{h\rightarrow0} \frac{y(x+h)-y(x)}{h}$$
Now, for simplicity denote the same limit as $y'(x)$. We want to express the above equality as an equality between the finite difference $y(x+h)-y(x)$ and the derivative, and not between it and the limit of the incremental quotient. The way to do this is to write $$y'(x)=\frac{y(x+h)-y(x)}{h}+r_x(h)$$
Where $r_x(h)$ is some remainder function that goes to zero as $h$ goes to zero. Intuitively, all you're saying is that if two quantities approach each other, then their difference, $r_x(h)$ must approach zero.
Now we define $o(h)$ to be any function that goes to zero faster than linearly (i.e $o(h)/h\rightarrow 0$), and thus we may write $$y(x+h)-y(x)=y'(x)h+o(h)$$
It is also costumary to write this as $$\Delta y[h]=dy[h]+o(h)$$
Where $\Delta y[h]=y(x+h)-y(x)$ and we omit the point $x$ from the notation, since it is being fixed. We'll go over in a moment why this notation is good. But first, notice that the definition of $o(h)$ will absorb any quadratic terms or higher order terms since these are functions that go to zero faster than $h$. This justifies the (physicist-like) intuition that a differential is a "small difference" and that a "small difference" means ignoring quadratic terms and beyong (likely the reason why you came up with $(\Delta x)^2\rightarrow 0$), because all like terms won't end up in $dy$ but rather in $o(h)$ in the above expression.
Now, why is this good notation? In one dimension, it's a bit hard to justify, but it has to do with linear algebra.
First, we remark that the expression $dy[h]=y'(x)h$ is a linear function of $h$. Then we could write $dy=y'(x)\cdot\_$, meaning, $dy(x)$ is the linear map that gives multiplication by the number $y'(x)$. But there's another way to write it. Let $x$ be the identity map $x \mapsto x$, with a slight abuse of notation. Then, notice that $$x+h-x=h=h+0$$ and $0=o(h)$. Hence the derivative of $x$ is $1$ (unsurprisingly) and we may write $dx[h]=1\cdot\_$.
But then, with all this notation, we can write $dy[h]=y'(x)\cdot h=y'(x)\cdot 1\cdot h=y'(x)dx[h]$, or more compactly, $dy=y'(x)dx$.
This is the correct definition of the differential. To state it clearly, the differential of a function $y$ is the function that, to any point $x$, gives the linear map $dy(x)=y'(x)dx$, which applied to an increment $h$, based on $x$, gives (up to quadratic order) the variation of $y$ on that increment.
To summarize,
- the differential is not a small difference
- quadratic terms get ignored because the differential is a linear approximation map
- The two definitions (derivative and differential) are equivalent in the sence that the action of the differential at the point $x$ is to multiply the increment by the derivative of $y$ at $x$; hence there is a one to one correspondence between $dy(x)$ and $y'(x)$
