How does the left side sees one coil less than the right side?

I think one way to approach this is to consider the case that the bolded wire from $a$ to $b$ is just that - a wire from $a$ to $b$ that doesn't wind around the core multiple times. Let's say that this is the $N = 1$ case.

Now, consider the path from the top of $O_2$ clockwise to $a$ then 'round the loop to $b$ and then to the bottom of $O_2$. This loop is not threaded by the flux $\phi(t)$ (the closed path does not enclose the core section).

Now, consider the path from the top of $O_1$ counterclockwise to $a$ then 'round the loop to $b$ and then to the bottom of $O_1$. This loop is threaded by the flux $\phi(t)$ (the closed path does enclose the core section).

It follows that $u' = 1\cdot\frac{d\phi}{dt}$ (flux threads one loop) and $u'' = 0\cdot\frac{d\phi}{dt}$ (flux threads zero loops).

From this, deduce that the general result for $N > 1$.

How does the left side sees one coil less than the right side?

I think one way to approach this is to consider the case that the bolded wire from $a$ to $b$ is just that - a wire from $a$ to $b$ that doesn't wind around the core multiple times. Let's say that this is the $N = 1$ case.

Now, consider the path from the top of $O_2$ clockwise to $a$ then 'round the core to $b$ and then to the bottom of $O_2$. This loop is not threaded by the flux $\phi(t)$ (the closed path does not enclose the core section).

Now, consider the path from the top of $O_1$ counterclockwise to $a$ then 'round the core to $b$ and then to the bottom of $O_1$. This loop is threaded by the flux $\phi(t)$ (the closed path does enclose the core section).

It follows that $u' = 1\cdot\frac{d\phi}{dt}$ (flux threads one loop) and $u'' = 0\cdot\frac{d\phi}{dt}$ (flux threads zero loops).

From this, deduce that the general result for $N > 1$.