It seems that there is a action by which $SL_2(\mathbb{F}_p)$ and $GL_2(\mathbb{F}_p)$ permute the $p^2$ ordered tuples in $\mathbb{F}_p^2$. What is the map from the $2 \times 2$ matrices over $\mathbb{F}_p$ versions of the elements of these groups to the elements of $S_{p^2}$ that this action picks out?
Similarly there seems to be an injective homomorphism map from $PSL_2(\mathbb{F}_p)$ to $S_{p+1}$ by virtue of the natural action on $\mathbb{F}_pP^1$ (the projective space obtained from $\mathbb{F}_p^2$) What is this mapping?
A matrix $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \in \text{GL}_2(\mathbb{F}_p)$ acts on a vector $v = \left[ \begin{array}{cc} v_0 \\ v_1 \end{array} \right] \in \mathbb{F}_p^2$ by matrix multiplication:
$$Av = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} v_0 \\ v_1 \end{array} \right] = \left[ \begin{array}{cc} a v_0 + b v_1 \\ c v_0 + d v_1 \end{array} \right].$$
This is far and away the best description of this action I can give. You can figure out everything you could possibly want to know about this action from linear algebra. It is even better than knowing a permutation representation because linear algebra is easier than set theory, even finite set theory.
For example, here is how you figure out the cycle decomposition of the permutation corresponding to $A$. It splits into cases depending on the conjugacy class of $A$ as follows.
Case: $A$ is diagonalizable over $\mathbb{F}_p$. Then up to conjugacy it is a diagonal matrix with diagonal entries $\lambda_0, \lambda_1 \in \mathbb{F}_p^{\times}$. Its order is the lcm of the orders $d_i$ of $\lambda_i \in \mathbb{F}_p^{\times}$, which can be any divisor of $p - 1$. The corresponding permutation of $\mathbb{F}_p^2$ fixes zero and splits the nonzero vectors into cycles as follows:
Case: $A$ is diagonalizable over $\mathbb{F}_{p^2}$ but not over $\mathbb{F}_p$. Then up to conjugacy its action on $\mathbb{F}_p^2$ is the action of multiplication by an element $\lambda \in \mathbb{F}_{p^2}^{\times}$ (one of the eigenvalues of $A$) on $\mathbb{F}_{p^2}$. Its order is the order $d$ of $\lambda$ in $\mathbb{F}_{p^2}^{\times}$, which can be any divisor of $p^2 - 1$. The corresponding permutation of $\mathbb{F}_{p^2}$ fixes zero, and the nonzero vectors form cycles of length $d$.
Case: $A$ is not diagonalizable even over $\mathbb{F}_{p^2}$. From the theory of rational canonical form we deduce that up to conjugacy $A$ is a Jordan block
$$A = \left[ \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right]$$
where $\lambda \in \mathbb{F}_p^{\times}$. Its order is $p$ times the order $d$ of $\lambda$, which can be any divisor of $p - 1$. The corresponding permutation of $\mathbb{F}_p^2$ fixes zero and splits the nonzero vectors into cycles as follows:
The analysis for the action on the projective line is similar. Writing a point on the projective line in projective coordinates $(v_0 : v_1)$, the action takes the form
$$A (v_0 : v_1) = (av_0 + bv_1 : cv_0 + dv_1)$$
or, thinking of an element of the projective line as a fraction $z = \frac{v_0}{v_1}$ (which may take the value $\infty$),
$$A z = \frac{az + b}{cz + d}.$$
Again you can figure out everything you could possibly want to know about this action from linear algebra as above. In the diagonalizable case, for example, projectively the action of a diagonal matrix with entries $d_0, d_1$ is multiplication by $\frac{d_0}{d_1}$, and hence its cycle decomposition is determined by the order of $\frac{d_0}{d_1}$. So things are even simpler.