The electric field lines shown in your diagram are for the electrostatic case of two oppositely charged circles.
When analyzing even the simplest direct current (DC) circuits, it is important to realize that the situation is non-electrostatic, since charges are moving through the conductor.
The essence of circuit theory can be understood with reference to Kirchoff's circuit laws, which were devised some years prior to Maxwell.
Of course, these laws are embodied in Maxwell's equations, as we can see from the Maxwell-Ampere law:
$\nabla\times\textbf{E}=\dfrac{\partial\textbf{B}}{\partial t}$
or in integral form,
$\oint \textbf{E}.d\textbf{l} = \iint \dfrac{\partial \textbf{B}}{\partial t}.d\textbf{S} $
Which, for electrostatic and magnetostatic (DC current) cases, become:
$\nabla\times\textbf{E}=0$
or in integral form,
$\oint \textbf{E}.d\textbf{l} = 0$
This means that the electric field is conservative in these cases. That is, we can define a scalar function (the electric potential, V), where:
$\textbf{E}=-\nabla V$
In the absence of any other forces or constraints, a small positive test charge will follow the path shown by the blue line, so the electric field lines indicate the path that a 'free' charge will take. Note that there are not normally any 'free' charges in the air, as the air is a dielectric (insulator).
Now suppose a lamp is connected to the terminals of the charged battery by two pieces of resistive wire along some arbitrary path. Assume also that the overall charge of the battery/wire/lamp system is neutral.
When the wire is connected to the charged battery terminals, the electrons move along the wires and through the lamp, although they experience some 'resistance' due to 'frictional' forces within these conductors which reduce the mobility of charge carriers. These frictional forces are generally proportional to the velocity of the charge carrier and result in heat dissipation in the wire and lamp, as well as light being emitted by the lamp filament.
Whilst the battery continues to pump charges, a charge gradient is maintained along the length of the wire, with higher concentration of net positive charges near the positive terminal, dropping to zero at some intermediate point along the wire and then decreasing further to net negative charge further along the wire toward the negative terminal. Where the wire 'doubles back' toward the positive terminal, a slight positive net charge density concentrates at the outer surface (furthest away from the positive terminal) whilst further along at the next bend, a slight negative net charge density concentrates close to the positive terminal, thereby maintaining an weak electric field along the length of the wire and allowing current to flow back to the battery.
Note that whist the current flows through the wire, the electric field outside the wire is distorted with respect to the electrostatic case, due to the charge concentration at the surface of the bends in the wire.
This is analogous to a river meandering down a gentle slope. Just as the flow of water carves out the path of a river, so too the flow of current through a wire creates an electric field. Under high current flows, the change in momentum of the water around a bend results in a higher thrust force and hence pressure on the outer bank of the river. As long as current flows through the wire, there is a slight but steady (monotone) drop in potential along the length of the wire from positive to negative, despite the wire doubling back towards the positive terminal, because the current in the wire alters the charge distribution and hence the electric field (forces) and potential gradient (pressure) along the wire.