= Solution
Introduce <conformal time>
$$
\eta=\int_0^t\frac{ds}{\cosh s}=2\arctan[\tanh(t/2)]=\arcsin(\tanh t),
\qquad -\frac\pi2<\eta<\frac\pi2.
$$
It satisfies $d\eta/dt=\operatorname{sech}t$ and $\cos\eta=\operatorname{sech}t$, so the metric becomes $ds^2=\sec^2\eta(-d\eta^2+d\xi^2)$. Choose <null coordinates>
$$
\boxed{u=\xi-\eta,\qquad v=\xi+\eta,\qquad
ds^2=\sec^2\!\left(\frac{v-u}{2}\right)du\,dv.}
$$
Thus $f(u,v)=\sec[(v-u)/2]$ is positive on $-\pi<v-u<\pi$. Choosing the two null coordinates with these orientations gives the requested positive coefficient of $du\,dv$.
There is a global-topology distinction in the source: unrestricted $\xi\in\mathbb R$ gives the <universal cover of two-dimensional de Sitter spacetime>. The usual <two-dimensional de Sitter spacetime> has $\xi$ periodic modulo $2\pi$. Indeed its standard embedding is $X^0=\sinh t$, $X^1=\cosh t\cos\xi$, $X^2=\cosh t\sin\xi$, and is unchanged by $\xi\mapsto\xi+2\pi$. The local metric and the null-coordinate calculation apply to either choice.
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