Assume we have a parametrized FO formula of this form:
$$\varphi(x_1,x_2, y_1, \dots ,y_m) := \xi(x_1,x_2) \land \psi(y_1,\dots,y_m)$$
We want to use as few additional bound (quantified by $\exists$ or $\forall$) variables as possible. The $\xi$ is not dependent on $y_1, \dots, y_m$ however it does require some additional bound variables.
For example $\varphi$ is a formula in the universe of natural numbers $\mathbb{N}$ and we have the natural $<$ relation. We want $\xi(x_1,x_2)$ to be satisfied if there are at least $m$ natural numbers inbetween $x_1$ an $x_2$. $\psi(y_1,\dots,y_m)$ is a formular that states whether $y_1 < \dots < y_m$.
An obvious solution is the following:
$$ \xi(x_1,x_2) := \exists z_1 \dots \exists z_m ( x_1 < z_1 \land z_1 < z_2 \land \dots \land z_{m-1} < z_{m} \land z_m < x_2)$$
$$ \psi(y_1,\dots,y_m) := y_1 < y_2 \land \dots \land y_{m-1} < y_{m} $$
$z_1, \dots, z_m$ are additional variables. $\varphi$ depends on $2+m$ parameters and requires $m$ additional variables. Hence, $\varphi \in \mathsf{FO}^{2m+2}$.
An alternative idea is to use $y_1,\dots,y_m$ as quantified variables:
$$ \xi(x_1,x_2) := \exists y_1 \dots \exists y_m ( x_1 < y_1 \land y_1 < y_2 \land \dots \land y_{m-1} < y_{m} \land y_m < x_2)$$
$$ \psi(y_1,\dots,y_m) := y_1 < y_2 \land \dots \land y_{m-1} < y_{m} $$
In this case $\varphi \in \mathsf{FO}^{m+2}$. But is it correct? Are $y_1,\dots,y_m$ the old parameters again after the brackets? I know we can just write $\varphi(x_1,x_2, y_1, \dots ,y_m) := \psi(y_1,\dots,y_m) \land\xi(x_1,x_2)$ to be on the safe side (the example is rather artificial in order to be simple). But anyway would it work?
In general (depending on signature and universe) the expressive power of $\mathsf{FO}^m$ is bigger then $\mathsf{FO}^n$ if $m>n$. So this construct would help to prove that a certain property is expressible with a certain number of variables.
In a formula such as $$ [(\exists x)R(x,z)] \land S(x) $$ the first occurrence of $x$ is bound and the second occurrence is free. The quantifier, in essence, temporarily wipes out any previous value that $x$ might have had. The formal way to see this is to examine the definition of what it means for an interpretation to satisfy a formula.
The order of formulas is not important, however. For any two formulas $\phi$ and $\psi$, $\phi \land \psi$ is equivalent in every logical way to $\psi \land \phi$. So if there was any issue with one of the, there would also be an issue with the other. Formulas are written from left to right, but the definition of what it means to satisfy a formula does not depend in any way on the order in which conjuncts are written.