Willn't the molecule collide with the opposite wall during that time-interval? It is moving to the left after being rebounded from the piston. So,willn't during the one-round trip, the molecule collide with the wall on the opposite side?
If I am right that $L$ is the cylinder lenght, then $\Delta t$ is the time from a molecule leaves the piston to it reaches it again. And we consider only one molecule know.
So the answer is yes. The molecule will hit the opposite wall and bounce back.
Assuming the molecule has horizontal speed $v_x$ (for more molecules $v_x$ is thus considered an average speed), during $\Delta t$ the molecule is sure to hit the piston exactly once. That might be the point of the text.
And why does the author want to exclude the acceleration imparted by the opposite wall??
Actually he is not "totally" excluding it - the rebouncing is necessary for the setup to make sense. If the opposite wall was not changing the direction of the molecule (rebouncing in an assumably elastic collision), which means changing the horizontal speed from $-v_x$ to $+v_x$, the piston would not necessarily be hit by the molecule during $\Delta t$.
He is just not including it in his calculations. To find pressure you don't need to consider more than one surface. By including as little as possible the model is kept simple.
If he is doing so, how can he measure the total pressure as he is ignoring the force of the molecule on the opposite wall??
Remember that pressure is per area, $p=F/A$. So
ignoring the force on the opposite wall is irrelevant. You can say, you only need to consider one square meter - or just one surface of which you know the exact area. In this case the piston.
He knows that exactly one molecule hits the piston during $\Delta t$ (that is, it is being accelerated by having the direction changed, $+v_x$ to $-v_x$). If you had $n$ molecules you then know that exactly $n$ collisions will happen during $\Delta t$.
He knows the mass $m$ (I assume he knows what gas molecule it is).
So at some point during $\Delta t$ the total mass $M$ (or $M=n*m$ for many molecules) is being accelerated by the piston area $A$. That force would be $F=Ma$ (or $F=Ma=nma$ for more molecules).
The pressure is just $p=F/A$ (or $p=F/A=nma/A$ for more molecules). It is like an average force per square meter during that time $\Delta t$.
The speed $v_x$ (which would be average speed for more molecules) depends on temperature e.g. If that $v_x$ is known, he can find the acceleration also and find the pressure.