Other answers have already addressed the relationship between General Relativity and the Minkowski metric, but it seems you are most interested in getting from the Minkowski metric to the Lorentz transformation. So let's do that.
Given a set of coordinates in which the metric takes the standard Minkowski form
$ds^2 = dt^2 - dx^2 - dy^2 - dz^2$
you want to find another set of coordinates in which the metric also takes the same form
$ds^2 = d \bar{t}^2 - d \bar{x}^2 - d \bar{y}^2 - d \bar{z}^2$
Consider a linear boost along the x axis. We want to choose coordinates such that $\bar{x} = 0$ where $x - vt = 0$; the general solution is:
$\bar{x} = \gamma(x - vt)$
$\bar{t} = at - bx$
where $\gamma$, $a$, and $b$ are unknowns.
Then
$dt^2 - dx^2 = d\bar{t}^2 - d\bar{x}^2$
$ = (a dt - b dx)^2 - \gamma^2 (dx - v dt)^2$
$ = (a^2 - \gamma^2 v^2) dt^2 - (\gamma^2 - b^2) dx^2 + (\gamma^2 v - ab) dx dt$
so
$a^2 - \gamma^2 v^2 = 1$
$\gamma^2 - b^2 = 1$
$\gamma^2 v - ab = 0$
so
$a = \sqrt{1 + \gamma^2 v^2}$
$b = \sqrt{\gamma^2 - 1}$
so
$\gamma^2 v = \sqrt{\gamma^2 v^2 + 1} \sqrt{\gamma^2 - 1}$
$\implies \gamma^4 v^2 = (\gamma^2 v^2 + 1)(\gamma^2 - 1)
= \gamma^4 v^2 + \gamma^2 - \gamma^2 v^2 - 1$
$\implies \gamma^2(1 - v^2) = 1$
$\implies \gamma = \sqrt{\frac{1}{1-v^2}}$
Calculating a and b (exercise left to the reader) gives $a = \gamma$ and $b = \gamma v$, so that
$\bar{t} = \gamma(t - vx)$
completing the Lorentz transformation as expected.