What is the magnetomotive force exactly? What is the magnetomotive force defined in terms of the variables of and derived from the Maxwell's equations? The emphasis is on the mathematical derivation.

The answer by @NicolasSchmid is wrong as explained by my comments below that answer.
 A: The magnetomotive force is the magnetic  analogous to the electromotive force. For electric current. You might know that the sum of all the electric currents through a surface is
$$\unicode{x222F}_A \vec{J} \cdot \vec{dA} = 0 $$
There is an equation that looks pretty similar to it in the Maxwell's equations $\unicode{x222F}_A \vec{B} \cdot \vec{dA} = 0$
So in this analogy, the current density $\vec{J}$ behaves the same as the magnetic field $\vec{B}$. So since the electric current is given by the equation $I = \iint_A \vec{J} \cdot \vec{dA}$ it seems natural to consider the magnetic flux $\phi_B = \iint_A \vec{B} \cdot \vec{dA}$ as the "magnetic current".
Yout can push the analogy further with the two formulas$$\vec{J} = \kappa \cdot \vec{E}, \qquad \vec{B} = \mu \cdot \vec{H}$$
So, the the permeability $\mu$ is analogous to the conductivity $\kappa$ for electric currents, and the electric field $E$ is analogous to the magnetic field strength $H$.
Since the the electrical voltage is defined as $U = \int_{\gamma} \vec{E} \cdot \vec{dr}$ and $H$ is analogous to $E$ it seems natural tol define the magnetic voltage $V_m= \int_{\gamma} \vec{H} \cdot \vec{dr}$. Of course, once you have the magnetic current and the magnetic voltage, you can define the magnetic resistance $R_m = \frac{V_m}{\phi_B}$.
The magnetomotive force has the ability to induce a magnetic flux. That flux will be given by $\phi_B = \frac{magnetomotive force}{R_m}$  so you can think about it as a magnetic voltage source.
For example if you have a coil with $N$ windings and a current $I$ flowing through, you know from Ampere's law that the integral over a closed loop that goes through the coil $\oint_\gamma \vec{H} \cdot \vec{dr} = N \cdot I$.
If you have a resistor in serie to a voltage source and you evaluate the closed loop intergral $\oint_\gamma \vec{E} \cdot \vec{dr} - \epsilon_{mf} = 0 \Longrightarrow \oint_\gamma \vec{E} \cdot \vec{dr} = \epsilon_{mf}$ So you can see by comparing these two equations that in this case the magnetomotive would be $N \cdot I$.
