= Solution
Under $y=e^{-2x}v$, the left <Robin boundary condition> becomes $v'(0)-v(0)=0$ and the right <Dirichlet boundary condition> becomes $v(1)=0$. Choose
$$
v_L(x)=\cos(\pi x)+\frac1\pi\sin(\pi x),\qquad v_R(x)=\sin(\pi x).
$$
Their <Wronskian> $v_Lv_R'-v_L'v_R$ is the constant $\pi$. For $y_L=e^{-2x}v_L$ and $y_R=e^{-2x}v_R$, it is $W_y(x)=\pi e^{-4x}$. The coefficient enforcing the derivative jump four is $4/W_y(\xi)$. Writing $x_< =\min(x,\xi)$ and $x_>=\max(x,\xi)$, the <Green's function> is
$$
\boxed{g(x,\xi)=\frac4\pi e^{2\xi-2x}\left[\cos(\pi x_<)+\frac1\pi\sin(\pi x_<)\right]\sin(\pi x_>).}
$$
The endpoint conditions, continuity and derivative jump follow directly from this <Wronskian> construction. Its exponential factor also shows why the kernel need not be symmetric in the unweighted inner product.
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