Here is an "intuitive" explanation. I will work on $S^1=[0,1]$ (identifying $0$ and $1$) and denote $e(x)=e^{2\pi ix}$. I will also denote $S_N=\{0,\frac 1N,\ldots,\frac{N-1}N\}\subset S^1$ and assume for convenience that $N$ is odd. Let $f\in C^{\infty}(S^1)$. Set $f_N=\sum_{n=-(N-1)/2}^{(N-1)/2}\hat f(n)e(nx)$. Then $f_N$ coincides with $f$ on $S_N$ by the Discrete Fourier Transform. Denote also $f_{N,m}=\sum_{n=-m}^m\hat f(n)e(nx)$.
There are two key observation to make:
* While $f_N$ usually swings wildly outside of $S_N$, if $m$ is small compared to $N$ then $f_{N,m}$ does not, i.e. its derivative is $o(N)$.
* $\hat f(n)$ decays very fast with $n$ (faster than any polynomial), so $f_{N,m}\approx f_N=f$ on $S_N$ and therefore $f_{N,m}\approx f$ on $S^1$, because both functions have derivative $o(N)$.

You need to do some calculations to make this rigorous, but they are straightforward .