Why is the Plane progressive wave equation $y= a\sin (kx-wt)$ for positive direction of x-axis?

Let's take the argument of the function i.e, $kx-\omega t$. The argument of the function should remain constant,(equivalently the phase must remain constant)for a particular section of the wave. \begin{equation} kx-\omega t=\lambda \end{equation} where $\lambda$ is a constant. Differentiating both sides we get, \begin{equation} k\frac{dx}{dt}=\omega \end{equation} which is positive, thus this represents a wave travelling in positive direction. Similarly for the wave travelling in negative direction.

This can be illustrated by plotting also, the blue wave represents wave at $t=0$ and the orange wave represents wave at a later time, As you can see, the wave has moved in left direction for the first figure for the equation $y=a\sin(kx+\omega t)$ and in the right for the second i.e for $y=a\sin(kx-\omega t)$ case.Consider the blue wave below at some $x$, now if you want to make the wave move, then you need the same $y$ at some other $x$ and $t$, thus for the first case,it happens when $x$ becomes less positive, or moves to the left, implying that the wave has traveled in negative direction. wave travelling in the negative direction wave travelling in the positive direction

Suppose an x-axis which points to the right.
A wave produced by a source travels to the right.

The displacement due to the wave at position $x$ at a time $t$ is $y = f(kx \pm \omega t)$ where $f(kx \pm \omega t)$ is a function which satisfies the wave equation.
It might help your visualisation if you think of that displacement corresponding to a peak.

Since the wave and hence the "information" is travelling to the right at some later time $t + \Delta t$ a particle at $x + \Delta x$ will have the same displacement $y$.
In other words the peak has travelled a distance $\Delta x$ in a time $\Delta t$ with both $\Delta x$ and $\Delta t$ both positive.

So $y = f(kx \pm \omega t) = f(k(x + \Delta x) \pm \omega (t + \Delta t))$

which implies $kx \pm \omega t = k(x + \Delta x) \pm \omega (t + \Delta t) \Rightarrow 0 = k\Delta x \pm \omega \Delta t$

Since in the case of a right travelling wave $k, \Delta x, \omega $ and $\Delta t$ are all positive the only way to satisfy this equation is to have $0 = k\Delta x - \omega \Delta t$ which implies that the original function was $y = f(kx - \omega t) $.

Out of this analysis you also get that $\frac {\Delta x }{\Delta t} = \frac \omega k $ which is the speed of the wave.

When the information is travelling to the left then either $-\Delta t$ and $+\Delta x$ or $+\Delta t$ and $-\Delta x$.

To satisfy $0 = k(-\Delta x) \pm \omega \Delta t$ or $0 = k\Delta x \pm \omega (-\Delta t)$ it must be that $y = f(kx + \omega t) $.

So if you want to watch the progress of a right travelling wave, ie follow a peak as time $t$ and position of peak $x$ increases you need to keep $kx - \omega t$ constant.