If my computations are correct, there is a root of the form $x=1+\sin^2(\frac\theta2)$ with:
$$\theta \in \left(\frac{k\pi}N,\frac{k\pi}N+\frac\pi{2N}\right) \text{ if }k < \frac{N}3-\frac12$$
$$\theta \in \left(\frac{k\pi}N-\frac{\pi}{2N},\frac{k\pi}N\right)\text{ if }k> \frac{N}3+\frac12.$$
I localize the zeros using another expression of $f(1+x)$.
For every $n \ge 1$, call $T_N$ the $n^{\rm th}$ Tchebycheff polynomial:
for all $\theta \in \mathbf{R}$, $$T_n(\cos\theta) = \cos(n\theta).$$
Set $c_k = \cos(k\pi/(2N))$ and $s_k = \sin(k\pi/(2N))$ for $0 \le k \le 2N$.
Then $1=c_0>\ldots>c_{2N}=-1$. The $c_k$ for $1 \le k \le 2N-1$ are
double roots of $T_{4N}-1$ whereas $-1$ and $1$ are simple roots.
The leading coefficient of $T_{4N}-1$ is $2^{4N-1}$. Thus
$$T_{4N}(X)-1 = 2^{4N-1}(X-1)(X+1) \prod_{k=1}^{2N-1}(X-c_k)^2.$$
Since $T_{4N}(X) = T_{2n}(T_2(X))$ and $c_{2N-k} = -c_k$ for $0 \le k \le 2N$,
we derive
$$T_{2N}(2X^2-1)-1 = 2^{4N-1}(X^2-1)X^2 \prod_{k=1}^{N-1}(X^2-c_k^2)^2.$$
Therefore,
$$T_{2N}(2X-1)-1 = 2^{4N-1}(X-1)X \prod_{k=1}^{N-1}(X-c_k^2)^2.$$
$$T_{2N}(2X-1)-1 = 2^{4N-1}\frac{X-1}{X} \prod_{k=1}^{N}(X-c_k^2)^2.$$
Taking logarithmic derivatives, we get
$$\frac{2T'_{2N}(2X-1)}{T_{2N}(2X-1)-1} = \frac{1}{X-1} - \frac{1}{X}
+ 2\sum_{k=1}^{N} \frac{1}{X-c_k^2}.$$
Replacing $X$ with $1-X$, we get
$$\frac{2T'_{2N}(1-2X)}{T_{2N}(1-2X)-1} = -\frac{1}{X} - \frac{1}{1-X}
+ 2\sum_{k=1}^{N} \frac{1}{s_k^2-X}.$$
$$\sum_{k=1}^{N} \frac{1}{s_k^2-X}
= \frac{T'_{2N}(1-2X)}{T_{2N}(1-2X)-1} + \frac{1}{2X(1-X)}.$$
Observe now that for every $k$,
$\sin^2(k\pi/N) = (2s_kc_k)^2 = 4s_k^2(1-s_k^2)$, so
$$\frac{1}{4}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= \frac{s_k^2}{s_k^2-X} - \frac{s_k^4}{s_k^2-X}.$$
$$\frac{1}{4}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= \Big(1 + \frac{X}{s_k^2-X}\Big) - \Big(s_k^2 + \frac{s_k^2X}{s_k^2-X}\Big).$$
$$\frac{1}{4}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= \Big(1 + \frac{X}{s_k^2-X}\Big)
- \Big(s_k^2 + X\Big(1 + \frac{X}{s_k^2-X}\Big)\Big).$$
$$\frac{1}{4}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= c_k^2 - X + \frac{X-X^2}{s_k^2-X}.$$
By summation over all $k \in \{1,\ldots,N\}$, since $c_N=0$
and by the equalities $c_{N-k}=s_k$ and $c_k^2+s_k^2=1$,
$$\sum_{k=1}^{N-1} c_k^2 = \sum_{k=1}^{N-1} s_k^2 = \frac{N-1}{2},$$
we get
$$\frac{1}{4} \sum_{k=1}^{N}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= \frac{N-1}{2} - NX + X(1-X) \sum_{k=1}^{N} \frac{1}{s_k^2-X}.$$
Putting thing together, we get
$$\frac{1}{4} \sum_{k=1}^{N}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= \frac{N-1}{2} - NX
+ X(1-X) \frac{T'_{2N}(1-2X)}{T_{2N}(1-2X)-1} + \frac{1}{2}.$$
$$1 - f(X+1) = \frac{1}{N} \sum_{k=1}^{N}\frac{\sin^2(k\pi/N)}{s_k^2-X}
= 2 - 4X + \frac{4}{N}X(1-X) \frac{T'_{2N}(1-2X)}{T_{2N}(1-2X)-1}.$$
$$f(X+1) = -1 +4X - \frac{4}{N}X(1-X) \frac{T'_{2N}(1-2X)}{T_{2N}(1-2X)-1}.$$
Let us evaluate at $X = (1-\cos\theta)/2 = \sin^2(\theta/2)$.
$$f(1+\sin^2(\theta/2)) = -1 + 2(1-\cos\theta) - \frac{1}{N}\sin^2\theta
\frac{T'_{2N}(\cos\theta)}{T_{2N}(\cos\theta)-1}.$$
By derivating $T_{2N}(\cos\theta) = \cos(2N\theta)$, we get
$\sin\theta \times T'_{2N}(\cos\theta) = 2N\sin(2N\theta)$. Hence
$$f(1+\sin^2(\theta/2)) = 1 - 2\cos\theta - 2\sin\theta
\frac{\sin(2N\theta)}{\cos(2N\theta)-1}.$$
$$f(1+\sin^2(\theta/2)) = 1 - 2\cos\theta + 2\sin\theta
\frac{\cos(N\theta)}{\sin(N\theta)}.$$
Now assume that $\theta = (k\pi+\epsilon)/N$ with $k \in \{1,\ldots,N-1\}$
and $|\epsilon|<\pi/2$.
Then
$$f(1+\sin^2(\theta/2)) = 1 - 2\cos\Big(\frac{k\pi+\epsilon}{N}\Big)
+ 2\sin\Big(\frac{k\pi+\epsilon}{N}\Big)
\frac{\cos\epsilon}{\sin\epsilon}.$$
Let us apply the intermediate value theorem.
$$\lim_{\epsilon \to 0+}f(1+\sin^2(\theta/2)) = +\infty.$$
$$\lim_{\epsilon \to 0-}f(1+\sin^2(\theta/2)) = -\infty.$$
$$\lim_{\epsilon \to \pi/2}f(1+\sin^2(\theta/2))
= 1 - 2\cos\Big(\frac{k\pi+\pi/2}{N}\Big).$$
$$\lim_{\epsilon \to -\pi/2}f(1+\sin^2(\theta/2))
= 1 - 2\cos\Big(\frac{k\pi-\pi/2}{N}\Big).$$
Since $k+\frac12 < \frac{N}3$ implies $2\cos((k\pi+\pi/2)/N)>1$, and
$k-\frac12 > \frac{N}3$ implies $2\cos((k\pi-\pi/2)/N)<1$, this gives the answer stated above.
One could refine the method and look at the sign of $f(1+\sin^2(\theta/2))$
when $\epsilon = \pm \pi/4$ for example.
