Electric field in a non-ideal MOSCAP
Your first equation for electric field is incorrect:

$$
E_{ox} = \frac{V_{fb}}{t_{ox}} = \frac{\phi_{ms}}{t_{ox}}-\frac{Q_f}{K_{ox}\epsilon_0}
$$

The electric field is the voltage across the oxide divided by the oxide thickness. \$V_{fb}\$ is not the voltage across the oxide at flatband. 

The correct electric field (at flatband) is:

$$
E_{ox}=\frac{Q_f}{K_{ox}\epsilon_0}
$$

To expand upon this, we assume there is no charge in the metal (normal assumption for a metal) or oxide (there could be, but there isn't in this question) and at flatband there is no charge in the semiconductor (by definition). The only place there can be charge in this structure is at the metal-oxide and metal-semiconductor junctions. Charge can move to the metal-oxide junction freely, but there are limitations on the oxide-semiconductor junction. In order to put charge at this junction you either need interface charge, which we know to be \$Q_f\$, or you have to put the semiconductor into accumulation or depletion mode. But we are at flatband, so we aren't in accumulation or depletion. Therefore, there is no additional charge at the oxide-semiconductor interface, meaning the voltage across the oxide is not a function of \$\phi_{ms}\$ but only a function of \$Q_f\$. From this you can integrate the charge to get electric field, which is therefore only a function of \$Q_f\$.