Why should we use the fundamental definition of derivative while checking differentiability?

The issue here is that the derivative $y'(1)$ is not defined. I'll start with the definition and then discuss why the differentiation rules don't apply at $x = 1$. First, for the derivative to exist at $x = 1$, then by definition,the following limit has to exist: $$\lim_{h\to 0}\frac{y(1+h) - y(1)}{h} \tag{1}.$$ Now, in order for $(1)$ to exist the left and right limits have to exist and agree. So, we can look at the left and right derivatives:

Right Derivative: $$y'_{+}(1) = \lim_{h\to 0^{+}}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0^{+}}\frac{2(1+h) - 2}{h} = \lim_{h\to 0^{+}}\frac{2h}{h} = 2$$ There's no trouble there, however we run into problems with the left derivative:

Left Derivative: \begin{align} y'_{-}(1)&=\lim_{h\to 0^{-}}\frac{y(1+h) - y(1)}{h}\\ &= \lim_{h\to 0^{-}}\frac{(1+h)^2 - 2}{h}\\ &= \lim_{h\to 0^{-}}\frac{1 +2h+h^2-2}{h}\\ &= \lim_{h\to 0^{-}}\frac{h^2+2h-1}{h}\\ &= \lim_{h\to 0^{-}}\left(h + 2 - \frac{1}{h}\right)\\ &= \infty \end{align}

So, since the left and right derivatives don't agree, we can say that $y$ is not differentiable at $y = 1$. As has been pointed out in the comments we could also have concluded that from continuity, because if $y'(1)$ had existed, then $y$ would need to be continuous at $x = 1$, and because it is not we can immediately rule out differentiability there.

Now, to understand why it doesn't work to take the derivatives of $2x$ and $x^2$ and take a limit, we start by noting that we can certainly write $$y'(x) = \begin{cases} 2x,&-\infty<x<1\\ 2, &1<x<\infty \end{cases}$$

However, if we were to write $$y'(1) = \lim_{x\to 1}y'(x)$$ we would be making two unfounded assumptions:

First, that $y$ is actually differentiable at $x = 1$, and so writing $y'(1)$ makes sense, and second that $y'$ is continuous. We don't know a priori that either of those things are true, and in fact as previously mentioned, we know that $y'(1)$ can't exist because $y$ isn't continuous at $x = 1$.

To give a bit more of an intuitive explanation, remember that the derivative at a point relies on the value of the function at that point. So, when we use $2x$ for the derivative on the left, instead of using $y(1) = 2$, which lies on the right piece, we're really using $\lim_{x\to 1^{-}}y(x) = 1$ on the left piece.