Your teacher's description is not bad. The phrase about mutual pushing is vague. I'm not sure if he or she means there is pushing to get things started, or pushing to maintain current, or something else. I think it might be fair to say that mutual pushing establishes the charge distribution needed to maintain the current, which I'm about to describe.
Your picture is pretty good, too. Once the current is established, charges accumulate on the surface of the wire in such a way that the surface charge density is positive near the positive battery terminal, negative near the negative battery terminal, and passes through zero somewhere in the middle. The result of this gradient of surface charge is to induce a uniform electric field inside the wire, much as you have drawn. It's this field that applies force to the charge carriers in the wire. You might argue that the charge carriers will accelerate without bound (Newton's second law), but no, each carrier will eventually collide with an impurity or defect and stop (or deflect, or turn back) the carrier, thus limiting the speed. A thermal vibration can do the same. Higher resistance materials have more impurities and defects, and thus lower average carrier speed. Raising the temperature of the material increases the number of thermal vibrations and also raises the resistance. This effect is prominent in a light bulb.
Classically, electrons do move in a conductor that is passing direct current – but much more slowly than you might think. Let's break this down:
Current in a wire is defined as the amount of charge that passes through a cross-section of that wire in a single second. By this definition alone, it is clear that a current relies on the motion of some charged particle. It is possible that there could be a system where electrons transfer energy to each other, but in classical terms this would not be considered a "current." However, as I mentioned before, electrons actually move pretty slowly, even in very high-power currents. This might be what you're thinking of – how even very slow-moving electrons transfer a lot of power.
As a matter of interest, let's look at exactly how quickly electrons move. We need a common identity, $I = qnA\overline{v},$ where $q$ is the charge of the charge carrier, $n$ is the number of those particles per unit volume, $A$ is the cross-sectional area of the wire, and $\overline{v}$ is the average speed of these particles. This identity is fairly simple to derive – $q n$ is the charge density per unit volume, and $A \overline{v}$ is the average volume of particles that passes through a cross-section of the wire in a given second, so $q n A \overline{v}$ is the total charge that passes through a cross-section of a wire in a given second, or equivalently the current $I$.
Now, suppose we have a copper wire. Let's make it pretty thick – say, 1cm in diameter. Classically, a current in a copper wire is transmitted by electrons. So the charge of the charge carrier is $q = e = 1.6 \cdot 10^{-19}$ C.
To find $n$, we note that, from Wikipedia, "Copper has a density of $8.94$ g/cm$^3$, and an atomic weight of $63.546$ g/mol, so there are $140685.5$ mol/m$^3$. In 1 mole of any element there are $6.02\cdot 10^{23}$ atoms (Avogadro's constant). Therefore in $1$ m$^3$ of copper there are about $8.5 \cdot 10^{28}$ atoms ($6.02 \cdot 10^{23} \cdot 140685.5$ mol/m$^3$). Copper has one free electron per atom, so $n$ is equal to $8.5 \cdot 10^{28}$ electrons per m$^3$."
For $A$, our wire is circular and has diameter $1 cm$, so its cross-sectional area in square meters is $A = \pi r^2 = \pi (0.5 \cdot 10^{-2})^2 = 7.85 \cdot 10^{-5}$ m$^2$
Now let's suppose we have a current of $1$ ampere – a fairly strong current by most standards. The velocity of the moving electrons in the wire is
$$
\overline{v} = \frac{I}{q n A} = \frac{1 \text{ C/s}}{1.6 \cdot 10^{-19} \text{ C} \cdot 8.5 \cdot 10^{28} \text{ m}^{-3} \cdot 7.85 \cdot 10^{-5} \text{ m}^2} \approx 9.37 \cdot 10^{-7} \text{m/s}.
$$
So a centimeter-thick copper wire carrying an ampere of current only requires its electrons to move $9.37 \cdot 10^{-7}$ m/s on average. That's very slow! You'll note from the relation $\overline{v} = I / q n A$ that the thicker the wire becomes, the smaller the velocity of the electrons is – that is, as $A$ grows larger, $\overline{v}$ grows smaller. Obviously, the wider a wire is, the more electrons can move through it. The more electrons moving through a wire, the slower they have to go to move the same total net charge.
The point is, power transferred in a wire is a result of massive numbers of electrons moving very, very slowly. They are moving, though.
Of course, this all only holds for direct current – as some other users have mentioned, alternating current is another common current in which electrons move back and forth with some predetermined frequency. In this case, the current is constantly switching directions, hence the name "alternating." This also could be what you're thinking of – as the electrons are moving back and forth but staying in essentially the same place. It is not the individual electrons that are transferring energy to each other, however, but the electric field pervading the wire which is constantly switching direction and forcing the electrons in the wire to change their direction of motion.
Best Answer
One Coulomb is defined as the charge transported by a current of 1 Ampere during 1 s. The Coulomb has nothing to do with the electric field generated by electrons.
As one electron has the negative charge of $1.602·10^{-19}C$, this means that 1 Coulomb has $6.24·10^{18}$ absolute values of electron, i.e. elementary charges. The Coulomb has thus a $6.24$ times a billion billion elementary charges