= Solution
Choose a left-end solution $u(x)=e^{-x}\sin x$ and a right-end solution $v(x)=e^{-x}(\sin x+\cos x)$. They satisfy $u(0)=0$ and $v'(2\pi)=0$. Their <Wronskian>, with the order used here, is
$$
W(x)=u(x)v'(x)-u'(x)v(x)=-e^{-2x}.
$$
The <Green's function> is $u(x_<)v(x_>)/W(\xi)$, where $x_<=\min(x,\xi)$ and $x_>=\max(x,\xi)$. Explicitly,
$$
\boxed{g(x,\xi)=-e^{\xi-x}\begin{cases}
\sin x(\sin\xi+\cos\xi),&0\leq x\leq\xi,\\
\sin\xi(\sin x+\cos x),&\xi\leq x\leq2\pi.
\end{cases}}
$$
The two expressions agree at the source. Their <derivative> difference is $[u(\xi)v'(\xi)-u'(\xi)v(\xi)]/W(\xi)=1$, checking both normalization and sign. The first-order <derivative> term means the <kernel> need not be symmetric under exchange of $x$ and $\xi$.
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