Question: What is $f(x)$ and what is $f(y)$?
You are given (no need to check) that the function $G(x-y,t)$ defined by
$$G(x-y,t)=\frac{1}{\sqrt{4\pi c^2 t}}e^{-(x-y)^2/4c^2t}$$
satisfies the $1$-D heat equation
$$G_t(x-y,t)=c^2 G_{xx}(x-y,t), -\infty \lt x \lt \infty, t\gt 0$$ Show that $u(x,t)$ defined by $$u(x,t) = \int_{-\infty}^\infty G(x-y,t)f(y) dy$$
also satisfies the $1$-D heat equation as well as the initial condition $$\lim \limits_{t\to 0^+} u(x,t)=f(x)$$
I think perhaps $f(y)$ isn't relevant, since: $$u(x,t) = \int_{-\infty}^\infty G(x-y,t)f(y) dy=\int_{-\infty}^\infty 0f(y)\,dy=0$$ And therefore this integral does satisfy the heat equation.
Perhaps for $f(x)$ they are saying that for $u(x,t)=F(x)G(t)$ we have $u(x,0)=F(x)$ e.g. $G(0)=1$. I am not totally sure.
Consider the one dimensional heat equation $$ \left \{ \begin{array}{cc} u_t - c^2 u_{xx} = 0 \\ u(x,0) = f(x) \end{array} \right. $$ It can be shown by direct verification that $$u(x,t) = G * f $$ where $*$ is the convolution operator, i.e. what you have written above. $f(x)$ is the initial data specified. Why does this come about? If you take a Fourier transform and solve the 1st order equation, you'll obtain $$ \hat u( \xi ,t) = \hat f ( \xi ) \hat G (\xi,t) $$ This is why it's nice to specify the time $t=0$ with something.