Another question about derivaton of Lorentz Tranformations

This question follows the same ideia of another question of mine: Doubts on derivation of Lorentz Transformations

So, the problem here is the following phrase:

The remain transformation equations for $x'$ and $t'$,namely,

\begin{cases}\tag{1} x' = a_{11}x+a_{12}y+a_{13}z+a_{14}t\\ t' = a_{41}x+a_{42}y+a_{43}z+a_{44}t\\ \end{cases}

Let us look first at $t'-equation$. For reasons of symmetry,we assume that $t'$ does not depend on $y$ and $z$. Other wise,clocks placed symmetrically in the $y-z$ plane, about the $x-axis$ would appear to disagree as observed from $S'$, which would contradict the isotropy of space. Hence $a_{42}=a_{43}=0$

We know that a point havind $x' = 0$ appears to move in the direction of the positive $x-axis$ with speed $v$ so the statemant $x'=0$ must to be identical to the statement $x=vt$

First of all, I didn't grasp quite well the arguments of the two bolded statements, I mean, the connection between geometry and physics. Principally when he said: "We know that a point havind $x' = 0$ appears to move..." and "clocks placed symmetrically in the y−z plane, about the x−axis would appear to disagree as observed from $S′$"

However, for the first bolded statement I think that means that for we have $a_{42}=a_{43}=0$ then $t'=0$ and $x=0$, $t=0$. So:

$$t' = a_{41}x+a_{42}y+a_{43}z+a_{44}t$$ become

$$0 = a_{41}0+a_{42}y+a_{43}z+a_{44}0$$

then

$$0 = a_{42}y+a_{43}z$$

Therefore,

$$a_{42}=a_{43}=0$$

But,as well as I said, I didn't grasp the physical significance of this manipulation (if it is correct).