There is not enough information to answer all of your questions; it depends on the behaviour of $f$. Perhaps useful is the observation that:
$F(x) = \int_{-\infty}^x f(y)\ \mathrm dy, G(c) = \int_{\Bbb R}\int_{-\infty}^{x+c} f(x)f(y)\ \mathrm dy \mathrm dx$
Assuming a sufficiently smooth density $f$ is used, we also have:
$G'(c) = \int_{\Bbb R} f(x)f(x+c) \ \mathrm dx$
Since $f(x) \ge 0$, $G'(c) \ge 0$ for all $c$. Note that it is not even a priori clear that this integral will exist.
Upon differentiating this expression again we get a term $f'(x+c)$ which can behave in arbitrary ways. Noteworthy is that since $0 \le F(x) \le 1$ for all $x$, we must have $0 \le G(c) \le 1$ for all $c \in \Bbb R$.
Now let us look at the domain of the double integration for $G(c)$. We have that it is:
$\{(x,y) \in \Bbb R^2: y \le x + c\} = \{(x,y) \in \Bbb R^2: x \ge y - c\}$
Thus we can rewrite the integral to:
$\int_{\Bbb R} \int_{y - c}^\infty f(x)f(y) \ \mathrm dx \mathrm dy$
but since $x,y$ are dummy variables, this is the same as:
$\int_{\Bbb R} \int_{x - c}^\infty f(x)f(y) \ \mathrm dy \mathrm dx$
If $c = 0$, we obtain:
$2 G(0) = \int_{\Bbb R} \int_{-\infty}^x f(x)f(y) \ \mathrm dy \mathrm dx + \int_{\Bbb R} \int_x^\infty f(x)f(y) \ \mathrm dy \mathrm dx = \int_{\Bbb R} \int_{\Bbb R} f(x)f(y) \ \mathrm dy\mathrm dx = 1$