There are $2$ copies of the trivial representation, and $p-1$ other representations. The two trivials are spanned by the characteristic function of $(0,0)$, and by the function which is $1$ on all the nonzero vectors and $0$ on $(0,0)$. The other representations are indexed by the characters of $\mathbb{F}_p^*$. Namely, for any character $\chi$, consider those functions $f:\mathbb{F}_p^2 \to \mathbb{C}$ such that $f(g \cdot v)=\chi(g) f(v)$ for $g \in \mathbb{F}_p^*$. This is a subrepresentation. It is irreducible except when $\chi$ is trivial, in which case the two trivials split off as summands.

**How to do this computation**: Let $V$ be a representation of $G$ over $\mathbb{C}$, and let $V:=\bigoplus W_{\chi} \otimes V_{\chi}$ be the decomposition into isotypic components, where $V_{\chi}$ are representatives for the isomorphism classes of $G$-irreps. Then
$$\mathrm{Hom}_G(V,V) = \bigoplus W_{\chi} \otimes W_{\chi}. \quad (*)$$
This formula is particularly useful when $V$ is a permutation representation $\mathbb{C}^X$, as then $\mathrm{Hom}_{\mathbb{C}}(V, V)$ is $\mathbb{C}^{X \times X}$ and the subalgebra of $G$-equivariant Homs is spanned by the $G$ orbits in $X \times X$. So the left hand side of $(*)$ has dimension equal to the number of $G$ orbits in $X \times X$. (In particular, we obtain the corollary mentioned by Qiaochu in a now-deleted answer: if the action of $G$ on $X \times X$ is doubly transitive, then $V$ is a trivial representation and one other irrep.)

In this case, there are $(p-1)+3+1$ orbits in $X \times X$. For every $g \in \mathbb{F}_p^*$, the set $\{ (v,w) : v=gw,\ v,w \neq 0 \}$ is an orbit. Also, we have three orbits by imposing that $v$, $w$ or both, be zero. Finally, there is the large orbit where $v$ and $w$ are linearly independent.

If we have already guessed the decomposition into irreducibles, as above, then we can confirm it by checking that the dimensions of both sides of $(*)$ match. In this case, that says $(p-1)+3+1=(p-1)+2^2$, which is true.

If we didn't already know the answer, we could deduce it from $(*)$. Let $R$ be the ring in $(*)$. We can describe $R$ as the vector space of $G$-invariant functions on $X \times X$, with the convolution product $(f*g)(u,w)=\sum_{v \in X} f(u,v) g(v,w)$. We must find the decomposition of $R$ into simple summands, and decompose $V$ as an $R$-module. The $R$-submodules of $V$ are the isotypic components for the $G$-action.

In this case, let $\Omega_g$ be the orbit $\{ (v,w) : v=gw,\ v,w \neq 0 \}$. It is easy to check that the $\Omega_g$ span a subalgebra isomorphic to the group algebra $\mathbb{C}[\mathbb{F}_p^*]$. This, of course, is isomorphic to $\bigoplus_{\chi} \mathbb{C}$, where the sum is over characters of $\mathbb{F}_p^*$. The idempotents for this decomposition project onto the corresponding $\chi$ representations above. The other semisimple summand of $R$ is a little harder to guess. Let $a_{00}$, $a_{10}$, $a_{01}$ and $a_{11}$ be the characteristic functions of the subsets of $X \times X$ where the two components are zero or nonzero according to whether the index is $0$ or $1$. Then the $a_{ij}$ span an algebra isomorphic to $\mathrm{Mat}_{2\times 2}(\mathbb{C})$, corresponding to the trivial summands.

You might enjoy reading the notes from the 2007 Quantum Gravity Seminar, which (despite its title) mostly focused on representation theory of finite groups. This is where I learned how to systematically use the ring $R$ to study permutation representations.

Algebra. The character of the permutation representation on $X$ is easy to compute. Using the orthogonality relations one gets the decomposition of the permutation character into irreducibles. With a little more work, one can get the decomposition on the level of actual representations. $\endgroup$ – Robin Chapman Jul 5 '10 at 6:39