What happens when a field turns on or off?
Quantum fields cannot be turned on or off. The field itself exists for all time and space. It is possible to excite various modes of a quantum field at various spacetime points. These field excitations are interpreted as particles. When no excitations are present (i.e. no particles are present) the quantum field is in the vacuum state. Particles do not act as a source for the fields; they are the excitations of the fields themselves. In an interacting theory, such as Quantum Electrodynamics, electrons, positrons and photons are allowed to interact. Contrary to the macroscopic picture, however, electrons do not act as a source of the electromagnetic field. Instead, the electron-positron spinor field interacts with the photon vector field.
Recall from the quantum harmonic oscillator the creation and annihilation operators $a^{\dagger}$ and $a$. The creation operator $a^{\dagger}$ has the property that when it acts on the $n$-th energy eigenstate, it yields the ($n+1$)-th energy eigenstate: \begin{equation} a^{\dagger}|{n}\rangle = \sqrt{n+1}|n+1\rangle, \end{equation} while the annihilation operator $a$ has the property that it lowers energy eigenstates: \begin{equation} a|n\rangle = \sqrt{n}|n-1\rangle. \end{equation} The vacuum state $|0\rangle$ is defined so that \begin{equation} a|0\rangle =0. \end{equation} In the canonical picture, quantum fields are defined in terms of creation and annihilation operators: \begin{equation} \phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}} \left[a_p(t)e^{-ipx}+a^{\dagger}_p(t)e^{ipx}\right], \end{equation} where \begin{equation} a_p^{\dagger}|0\rangle=\frac{1}{\sqrt{2\omega_p}}|p\rangle, \end{equation} with $\omega_p = \sqrt{\vec{p}^2+m^2}$. In other words, the quantum field $\phi(\vec{x})$ is an operator that acts on the vacuum and creates a particle at position $\vec{x}$.
Fundamental particles are modeled as point particles but this does not imply a "simple Gauss's Law" situation. The fact that Quantum Electrodynamics is mediated by the photon does not imply that there is any analogy between the behavior of macroscopic electric and magnetic fields and the quantized photons mediating the force. One striking difference is the fact that the Coulomb potential is not exactly $1/r$; higher order quantum effects contribute to logarithmic corrections to the strength of the electromagnetic force.