Four. (16 points) Given the quadratic function
$$
y=x^{2}-2(m-1) x+m^{2}-2 \text {. }
$$
(1) Prove: Regardless of the value of $m$, the vertex of the quadratic function's graph always lies on the same straight line, and find the function expression of this straight line;
(2) If the quadratic function's graph intercepts a segment of length 4 on the $x$-axis, find the function expression of this quadratic function.
(1) The vertex coordinates of the quadratic function's graph are $(m-1,2 m-3)$. Eliminating $m$ yields the function relationship as
$$
y=2 x-1 \text {. }
$$

Therefore, regardless of the value of $m$, the vertex of the quadratic function is always on this line.
(2) Suppose the graph of the quadratic function intersects the $x$-axis at points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$.

Given $\left|x_{2}-x_{1}\right|=4$, and using the relationship between roots and coefficients, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=2(m-1), \\
x_{1} x_{2}=m^{2}-2
\end{array}\right.
$$

Also, $\left(x_{2}-x_{1}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$, thus
$$
16=4(m-1)^{2}-4\left(m^{2}-2\right) \text {. }
$$

Solving for $m$ gives $m=-\frac{1}{2}$.
Therefore, the equation of this quadratic function is
$$
y=x^{2}+3 x-\frac{7}{4} \text {. }
$$