Given the heat equation on $x\in[0,1], t\geq 0$ with Dirichlet boundary condition $$\begin{cases} u_t = u_{xx}\\ u(x,0) = g(x)\\ u(0,t) = u(1,t) = 0 \end{cases}$$
From Fourier series, we have the Green's function $$G(x,y,t) = 2\sum_{n=1}^\infty e^{-n^2 \pi^2 t} \sin(n\pi x) \sin(n\pi y)$$
On the other hand, using fundemental solution $\Phi(x,t) = \frac{1}{2\sqrt{\pi t} }e^{-\frac{x^2}{4t}}$, we have another Green's function $$\tilde G(x,y,t) = \sum_{n\in \mathbb{Z}} \Phi(x-(2n+y), t) - \Phi(x-(2n-y),t)$$
I think they can not be equal because where the factor of $t$ appears in the sum, however the solution $u(x,t)$ can be given by both Green's functions in the same format... $$u(x,t) = \int_0^1 g(y) \tilde G(x,y,t) dy = \int_0^1 g(y) G(x,y,t) dy. $$
Here is the Poisson sum formula:
For a Schwartz function $f:\mathbb{R}\rightarrow \mathbb{C}$, the summation $$F(x)=\sum_{n \mathop \in \mathbb{Z}} f \left(x+{n}\right)$$ defines a $1$-periodic function on the unit interval $[0,1)\ni x$, and its fourier coefficient is $\hat f(m)$, so
$$\sum_{n \mathop \in \mathbb{Z}} f \left(x+{n}\right) = \sum_{m \mathop \in \mathbb{Z}} \hat f \left({m}\right)e^{im2\pi x}$$ when $x=0$ we get $$\sum_{n \mathop \in \mathbb{Z}} f \left({n}\right) = \sum_{m \mathop \in \mathbb{Z}} \hat f \left({m}\right)$$
This is not an answer to the question, but a comment too long to be edited in the comment section.
$$\begin{cases} u(x,0) = g(x)\\ u(0,t) = u(1,t) = 0 \end{cases} \quad\to\quad u(0,0)=g(0)=0\quad\text{and}\quad u(1,0)=g(1)=0$$
Expand the given function $g(x)$ in Fourier series on $0<x<1$ with $g(0)=g(1)=0$:
$$g(x)=\sum_{n=1}^\infty a_n\sin(n\pi x)$$ So, the coefficients $a_n$ are known.
Solving $u_t=u_{xx}$ thanks to the method of separation of variables leads to the solution : $$u(x,t)=\sum_{n=1}^\infty a_n\sin(n\pi x)e^{-n^2\pi^2t}$$