Note first that in your definition, assuming the usual that $A$ is an $n\times n$ matrix and $x$ an $n\times1$, $x^*Ax$ is a number. So your definition would be $ \|A\|=\max\{|x^*Ax|:\ x^*x=1\}. $ Now, this is the numerical radius of $A$ and not the operator norm (though they are equivalent as norms); the operator norm is defined as $\tag{1} \|A\|=\max\{\|Ax\|_2:\ x^*x=1\}=\max\{(x^*A^*Ax)^{1/2}:\ x^*x=1\}. $ The operator norm and the numerical radius agree for normal matrices, but not in general. For example, $\tag{2} A=\begin{bmatrix}0&1\\0&0\end{bmatrix}, $ satisfies $|x^*Ax|\leq1/2$ for all $x\in \mathbb C^2$, while $\|A\|=1$.
Regarding your question, it is not true that $ \max\{|x^*Ax|:\ x^*x=1\}=\max\{|y^*Ax|:\ x^*x=y^*y=1\} $ in general. This can be seen to fail for $A$ in the example $(2)$, where the number on the left is $1/2$, and the number on the right is $1$.
What is true is that $\tag{3} \|A\|=\max\{|y^*Ax|:\ x^*x=y^*y=1\}. $ The inequality $\geq$ is a direct consequence of Cauchy-Schwarz, and the inequality $\leq$ can be seen by considering the case where $y=Ax/\|Ax\|_2$.
Also true is that for selfadjoint $A$, $ \|A\|=\max\{|x^*Ax|:\ x^*x=1\}=\max\{|y^*Ax|:\ x^*x=y^*y=1\}. $ The proof of this is not hard, but not trivial either: below is the one from Conway: 
Finally, regarding your question 2). Think of the operator norm defined in $(1)$ and try to prove the triangle inequality. It is not impossible by any means, but not trivial either. On the other hand, it is trivial to prove it using $(3)$. The point is that in $(1)$ you are using a quadratic form, while in $(3)$ you are using a bilinear one.