Poynting vector in a simple DC circuit 
This image
I took from Wikipedia shows the directions of E(red) and B(green) and poynting vectors(blue) at different locations on the circuit. I don't seem to understand why the E fields(red) are pointing from up to down in that manner in between the circuit at plane P. Is it because the wire on the upper part is connected to the positive terminal and hence must E field lines must emanate from every part of the wire? 
 A: Suppose the wire was made out of copper with a diameter of 2 mm, and the + and - wires are distanced 10 cm from each other. Also let's assume the voltage was 1 V and the resistor was 100 Ohms, thus the current is 10 mA. We ignore the loss inside the battery and suppose it was an ideal voltage source.
The conductivity k of copper is ~ 6E7 S/m, thus the current density is ~ 3.18 kA/m². The strength of the electric field along the wire E = J/k is around 0.54 µV/m. The maximal field strength between the + and - wire is E = V/d = 1 V/10 cm 10 V/m for the shortest path, thus the electric field outside the wires (shown in this picture) is approx. 2 million times stronger than the field inside the wires (not shown). 
What about the magnetic field? Applying ampere's law to the infinite wire tells us that the magnetic field is proportional inside the wire, and falls off by ~ 1/r outside of the wire. The maximum magnetic field strength is thus closest to the surface of the wire, around 1.6 A/m. Only 5 cm away from the wire it drops off to 32 mA/m - 50 times weaker. 
Suppose the resistor is 6.8 mm long and has a diameter of 2.5 mm (taken from a data sheet). The maximum strength of the electric field E_max along the resistor will be approx. 1V/6.8mm = 150 V/m. The maximum poynting vector is E_max*H_max = 16 W/m². If you only take the poynting vector along the surface, the dissipated power is around 12.5 mW, for a more careful consideration you would need to perform integration over the entire volume of the divergence of the poyinting vector. 
A more simple calculation that comes fairly close is P = U * I = 10 mW. 
So yes - the power flows along the blue field lines, but most of it in very close proximity to the wire, so it does not really matter and we might just aswell say "power flows along the wires". 
A: There is a downward electric field in the P plane because the upper wire is at higher voltage than the lower wire.  So there is in effect more positive charge on the upper wire than the lower one.  
A: Short answer: There is a potential difference between wires, so there must be an electric field between those wires, going from positive to negative. The electric field inside the conductor is very, very weak, and ideally (in a perfect conductor) it doesn't exist at all.
More detail:
The concepts of voltage (electric potential) and electric field strength cannot be separated. One cannot have a potential difference between two points in space without an electric field existing between those two points. Voltage is a description of electric force between two points just the same as the electric field is a description of the electric force at a point. The difference is voltage is actually the electric potential energy per charge between two points i.e. how much energy is required to move a charge from one point to the other. This is directly related to the electric force on the charge, and thus the electric field in the region between the points. 
The electric field between two conductors with a potential difference (a voltage) between them will point from the positive wire to the negative wire, i.e. from the point of higher potential to the point of lower potential, because a charge in this area would be forced towards the lower potential. The closer two wires are with a given voltage between them, the stronger the electric field. The electric field will depend on the geometry of the circuit because of this. 
This is the reason for the familiar phenomenon of static shock occurring when you get closer to a doorknob. There's a potential difference between your hand and the doorknob, and the electric field gets stronger as you get closer to the doorknob, until you're so close it causes electrons to be stripped from air molecules (ionization of the air) and a rapid movement of charge. 
The electric field inside a conductor is only required for overcoming the resistance of a wire. Charge will not flow perfectly in a non-ideal conductor, so we need some non-zero force to keep charges flowing. In a superconductor, once a current is started, it can continue forever, with no electric field existing inside the conductor. There is only zero electric field in a conductor that is lossless. 
A: No. There is no electric field in the space enclosed by the current loop. The wires aren't oppositely charged, and the potential is the same everywhere inside the space enclosed. So there is no gradient of potential, and there is no electric field in the space between the wires. You aren't powering the circuit with a discharging capacitor. This misconception is everywhere, and it took me awhile to figure out. A lot of Physics people don't get batteries, and Chemistry-types don't explain Physics well. Batteries create an electric field in the circuit only. I also wonder about the field shown inside the battery in the first picture. Does it point in the wrong direction? The top of the battery isn't positively charged, and positive ions must flow up in the battery (or negative ions would flow down) across a porous material within the cell to ensure that no net charge accumulates anywhere. That suggests the electric field in the battery might point up, not down. But then again, if the field points up, then it would take work to move the positive ions up, and you could say that this is the work the battery does.
