Consider the two extremes: when $f = m$, the bonus is $0$; when $f=100$, the bonus is
$$\frac{0.1(100 - m)^2}{100 - m} = 0.1(100-m) = 10 - 0.1m.$$
At this latter extreme, your final grade will be
\begin{align}
0.60(100) + 0.20m + 0.20a + b &= 60 + 0.20m + 0.20a + 10 - 0.10m\\
&=70 + 0.10m +0.20a.
\end{align}
Notice that this is the average of (i) your normal weighted grade with (ii) the result of shifting all the weight of the midterm onto the final:
$$\frac12\left[(0.60(100)+0.20m+0.20a) + (0.80(100)+0.20a)\right] = 70+0.10m+0.20a$$
The curve connecting these two extremes is quadratic and has the same slope as the normal grade (without the bonus) at $f=m$. As a result, it only really makes a difference to those students who show a substantial improvement.
In the end, it's hard to see the point of such a scheme. It's unnecessarily complicated and doesn't really help students all that much. It would be far simpler (and slightly more generous) to give two marking schemes (in this case $60 / 20 / 20$ and $70 / 10 / 20$) and take the higher of the two.
Here's a graph.

From top to bottom:
Blue: 80% final (i.e., weight of midterm shifted to final)
Dotted red: my proposal for best of two marking schemes
Solid red: weighted grade + bonus
Purple: weighted grade without bonus
The vertical green line is the point where the final is better than the midterm (I arbitrarily chose $m=45$).
The curve at the very bottom shows the value of the bonus by itself.