Let $v_k(x) = x^k$, $k=0,...,4$. Then $v_k$ is a basis for $\mathbb{P}^4$. (To see this note that any quartic can be written in terms of $v_k$, and if $\sum \alpha_k v_k = 0$, then by differentiating and evaluating at $x=0$ we can see that $\alpha_k = 0$, hence they are linearly independent.)
By the same token, $v_k$, $k=0,1,2$ is a basis for $\mathbb{P}^2$. It follows that $\dim \mathbb{P}^2 = 3$, and since $\mathbb{P}^4 = \mathbb{P}^2 \oplus (\mathbb{P}^2)^\bot$, we see that $\dim (\mathbb{P}^2)^\bot = 2$.
We can find a basis for $(\mathbb{P}^2)^\bot$ by projecting the $v_k$ onto $(\mathbb{P}^2)^\bot$. Clearly $v_k$, $k=0,1,2$ will project to zero. So the only remaining basis elements that need to be projected are $v_3,v_4$.
Note in passing that $\langle v_j, v_k \rangle = \frac{1}{j+k+1}(1-(-1)^{j+k+1})$.
To compute the projection of $x$ onto $(\mathbb{P}^2)^\bot$, we need to determine $\alpha \in \mathbb{R}^3$ such that $\langle x-\sum_{k=0}^2 \alpha_k v_k, v_j \rangle = 0$ for $j=0,1,2$. This is just the linear system $\langle x, v_j \rangle = \langle \sum_{k=0}^2 \alpha_k v_k, v_j \rangle$, or $ \begin{bmatrix} \langle v_0, v_0 \rangle & \langle v_1, v_0 \rangle & \langle v_2, v_0 \rangle \\ \langle v_0, v_1 \rangle & \langle v_1, v_1 \rangle & \langle v_2, v_1 \rangle \\ \langle v_0, v_2 \rangle & \langle v_1, v_2 \rangle & \langle v_2, v_2 \rangle \end{bmatrix} \alpha = \begin{bmatrix} \langle x, v_0 \rangle \\ \langle x, v_1 \rangle \\ \langle x, v_2 \rangle \end{bmatrix} $ Grinding through the computations gives $\alpha = \frac{1}{5} (0,3,0)^T$ when $x=v_3$ and $\alpha = \frac{1}{35} (-3, 0, 30)^T$ when $x=v_4$.
Hence a basis for $(\mathbb{P}^2)^\bot$ is $x \mapsto x^3-\frac{3}{5}x$, $x \mapsto x^4+\frac{3}{35}-\frac{6}{7}x^2$.