Understanding this action from what you said requires a basic understanding of the objects of music theory as members of a set with some structure. I’ll try and give the briefest of explanations possible.
The set of pitch classes in (Western chromatic) harmony is a twelve-element set, which we will identify with the set of integers modulo 12. The natural cyclic ordering on the set of pitch classes agrees with the cyclic ordering on $C_{12} = \mathbb{Z}/12\mathbb{Z}$, so we will use it.
An interval is a pair of pitch classes $\{a,b\}$, and its quality is the quantity $q(a,b) = |a - b|$ mod 12. A minor third is an interval of quality 3, and a major third is an interval of quality 4. A triad is an ordered triple of pitch classes $(a,b,c)$. We are concerned with major and minor triads. A triad is major if $q(a,b) = 4$ and $q(b,c) = 3$, and a triad is minor if $q(a,b) = 3$ and $q(b,c) = 4$. (In other words, we always have $q(a,c) = 7$, and we call the triad major or minor depending on the quality of the interval $\{a,b\}$.) In music theory there are other triads, but for this answer I will use triad as a shorthand for major or minor triad.
The set of triads is a 24-element set, for a triad $(a,b,c)$ is completely characterized by its root note $a$ and its quality, major or minor. Thus it is convenient to parametrize triads as elements of the set $C_{12}\times C_2$. For some reason I would like to think of $C_2$ as the group of units of the ring $\mathbb{Z}$, forgive me. If $(a,b,c)$ is a major triad, it corresponds to the element $(a,+1) \in C_{12}\times C_2$. Likewise if $(a,b,c)$ is a minor triad, it corresponds to $(a,-1)$.
The action of transposition and inversion is simple to describe: consider the group of bijection of $C_{12}\times C_2$ generated by
$$\begin{cases}\iota\colon (a,\pm 1) \mapsto (-c,\mp 1) = (-a-7,\mp 1) \\ \tau\colon (a,\pm 1) \mapsto (a+1,\pm 1). \end{cases}$$ (Here and throughout addition and subtraction are mod 12.)
Note that $\iota$ has order $2$, $\tau$ has order $12$, and that $$\iota\tau\iota (a,\pm 1) = \iota\tau(-a-7,\mp 1) = \iota(-a-6,\mp 1) = (a-1,\pm 1) = \tau^{-1}(a,\pm 1),$$ so this defines an action of the dihedral group of order 24—which I am fond of writing as $D_{12}$ because of its action on a $12$-element set, forgive me—on the set of triads.
The “P/L/R” action is maybe slightly more involved.
The parallel major/minor of a triad $(a,\pm 1) \in C_{12}\times C_2$ is the triad $P(a,t) = (a,\mp 1)$. Thus $P^2 = 1$, and $P(0,3,7) = (0,4,7)$.
The leading-tone exchange of a major triad $(a,b,c)$ is the triad $L(a,b,c) = (b,c,a-1)$. Thus if $(a,b,c)$ is a major triad, $L(a,+1) = (b,-1)$. The leading-tone exchange of a minor triad $(a,b,c)$ is the triad $L(a,b,c) = (c+1,a,b)$. So if $(a,b,c)$ is a minor triad, then $L(a,-1) = (c+1,+1)$. One checks that $L^2 =1$.
The relative major of a minor triad $(a,b,c)$ is the triad $R(a,b,c) = (b,c,b+7)$—that is, if $(a,b,c)$ is a minor triad, $R(a,-1) = (b,+1)$. The relative minor of a major triad is the inverse operation: $R(a,b,c) = (b-7,a,b)$, so if $(a,b,c)$ is a major triad, $R(a,+1) = (b-7,-1)$. Similarly one checks that $R^2 = 1$.
I claim that the action of $RL$ on major triads is given by $RL = \tau^7$. (We are acting on the left.) Indeed, suppose $(a,b,c)$ is a major triad. Then $L(a,+1) = (b,-1)$, which is the minor triad $(b,c,b+7)$, so $RL(a,+1) = R(b,-1) = (c,+1)$.
On the other hand, the action of $LR$ on minor triads is given by $LR = \tau^7$ (so on minor triads we have $RL = (LR)^{-1} = \tau^{-7}$). Indeed, suppose $(a,b,c)$ is a minor triad. Then $R(a,-1) = (b,+1)$, which is the major triad $(b,c,b+7)$, so $LR(a,-1) = L(b,+1) = (c,-1)$. Thus in both cases, the quality is preserved, and the root note has shifted up $7$ pitch classes.
Since $7$ is relatively prime to $12$, $\langle R,L \rangle$ is a quotient of a group with presentation $\langle x,y \mid x^2 = y^2 = (xy)^{12} = 1\rangle$, and the latter is a presentation of $D_{12}$. I suppose one has to show that the action is faithful; it is, but I’ll leave that to you.
I claim further that $L$ and $R$ are sufficient to recover $P$. I don’t immediately see a slick way of doing this, so I’ll leave it to you.
To show that each centralizes the other, it suffices to show that $\tau R = R\tau$, $\tau L = L\tau$, $\iota R = R\iota$, and finally $\iota L = L\iota$. Let me maybe just talk through a tiny piece.
Suppose first $(a,b,c)$ is a minor triad. Then
$$\tau R (a,-1) = \tau(b,+1) = (b+1,+1) = R(a+1,-1) = R\tau(a,-1).$$ The argument for a major triad is analogous.
On the other hand, note that $L(a,-1) = (a-4,+1)$ and $L(a,+1) = (a+4,-1)$. Thus we have $$\iota L(a,-1) = \iota(a-4,+1) = (-a-3,-1) = L(-a-7,+1) = \iota(a,-1).$$ The argument for a major triad is analogous.
Of course, even buying the gaps I’ve left, this doesn’t show that these copies of $D_{12}$ are the full centralizer of each other in $S_{24}$, but hopefully you have some ideas about how to prove it.