First I want to introduce the notation I use:
So, for the Type I operation in wich we switch $l^{th}$ and $p^{th}$ rows, I denote it by $r_{lp}(A)$
So, if $1\le l\le p\le m$, then show that $r_{lp}(r_{pl}(A))=A$
Dem:
Let $A\in M_{(m,n)}(\mathbb{F})$ and $A_1=r_{lp}(A)$, defined as: $$ R_{i}(A_1) = \begin{cases} R_{i}(A), & \text{if $i\ne l,p$} \\[2ex] R_{p}(A), & \text{if $i=l$} \\[2ex] R_{l}(A), & \text{if $i=p$} \end{cases} $$ Then, let $A_2=r_{lp}(r_{pl}(A))$ or well, $A_2=r_{pl}(A_{1})$, defined as:
$$ R_{i}(A_2) = \begin{cases} R_{i}(A_1), & \text{if $i\ne l,p$} \\[2ex] R_{p}(A_1), & \text{if $i=p$} \\[2ex] R_{l}(A_1), & \text{if $i=l$} \end{cases} $$
So, we want to prove that $A_2=A=[\alpha_{ij}]$
For $i\ne l,p$ we have see that $R_i(A_2)=R_i(A_1)=R_i(A)$ wich we can interpret as $[ \alpha_{i1}\dots\alpha_{in}]$ for the $i^{th}$ row,
now for $i=l$, we have $R_l(A_2)=R_l(A_1)=R_l(A)$
Until here I seemed to lost the line of reasoning. The next should begin with the condition $l\le p$? Is the first part of the demonstration well defined?
To answer your question, it seems like the next case you want to check is the $i=p$ case.
Alternatively, you can view each row in $A$ as element of a set $X$ with size $m$, and switching two rows is the same as a 2-cycle (see here for more details) in $X$. By properties of the permutation group $S_X$, all 2-cycles have order 2; that is, applying a 2-cycle twice in a row is the same as doing nothing. This becomes more obvious in your case when you realize that $r_{lp} = r_{pl}$ (that is, they are the same operation).