What is the intuitution behind a wave with zero wavenumber?

In general, $n=0,1,2,...$. But if $n=0$, the field is zero, because the field is proportional to $sin(npi y/a)$. So the smallest value of $n$ for which the field can propagate is $n=1$.

In general, $k_x=0$ just means that the field is constant in the $x$ direction. A way think about this is that $k_x=2pi/lambda_x$, where $lambda_x$ is the wavelength in the $x$ direction. The limit of 0 wavenumber $k_xrightarrow 0$ corresponds to the limit of infinite wavelength $lambda_x rightarrow infty$. You can think of a constant as a wave with an infinite wavelength. Similarly, $|vec{k}|=0$ means that the field is a constant over all space.

I'm not quite sure why you have written the part of the field varying in the $x$ direction as $sin(k_x x - omega t)$, as opposed to $sin(k_x x-omega t +phi_0)$, for some initial phase $phi_0$. If this is allowed by the boundary conditions, then for $k_x=0$, we would have $omega=npi/a$, so the field would be vary in time but be constant in the $x$ direction, explicitly given by by $sin(- n pi/a + phi_0)$. For $phi_0=0$ (which is the expression you started with), the field would be identically zero.