A "Green's function", G(t, \tau), for a problem like this (the coefficient of the second derivative is 1) must satisfy several properties:
1) It must satisfy the homogenous d.e. in t for every \tau.
2) It must satisfy the boundary conditions.
3) It must be continuous.
4) The "jump" in the first derivative at t= \tau must be one.
Here, the general solution to the homogeneous equation is
e^t(C_1cos(kt)+ C_2sin(kt))
which means that the Green's function must be of the form
G(t, \tau)= \begin{pmatrix}e^t(Acos(kt)+ Bsin(kt) & 0\le t\le \tau \\ e^t(Ccos(kt)+ Dsin(kt) & \tau\le t\le 1\end{pmatrix}
To satisfy the first boundary condition, since 0\le\tau, we must have
G(0,\tau)= A= 0
To satisfy the second boundary conditon, we must have
G_t(0,\tau)= A+ B= 0
which, since A= 0, give B= 0. That is, G(t, \tau) is identically 0 for all 0\le t\le \tau.
But that does NOT say that G(t,\tau) must be 0 for \tau\le t\le 1.
The continuity condition gives that
\lim_{t\to\tau^+}G(t,\tau)= e^{\tau}(C cos(k\tau)+ Dsin(k\tau))= \lim_{t\to\tau^-} G(t,\tau)= 0
so we must have
e^{\tau}(C cos(k\tau)+ D sin(k\tau))= 0
which leads to
D= -\frac{cos(k\tau)}{sin(k\tau)}C
The "jump" condition on the derivative says that
\lim_{t\to\tau^-}G_t(t,\tau)- \lim_{t\to\tau^+}G_t(t,\tau)= 1
which gives
e^\tau(Ck cos(kt)+ Dk sin(kt))+ e^\tau(-Ck sin(kt)+ Dk cos(kt))= 1
those two equations give non-zero values for C and D.
