I am very new to K-theory, either topological or algebraic. Please bear with me if this sounds naively stupid.
After reading a bit of Wegge-Olsen, I have got the idea of how to calculate $K_0(C(S^2))$, by going through the song and dance of suspension and the cyclic six term exact sequence, etc. (I'll follow the subscript indices as in Wegge-Olsen, although most of the others use superscripts due to the contra/co-variance complication I guess.)
The rough picture I got was, either topologically we cover the two hemisphere with one patch for each and worry about the overlap on equator, or algebraicly write $C(S^2)$ as the one point compactification of $C_0(\mathbb{R}^2)$ and follow exactly
\begin{align} K_0(J) \xrightarrow{\iota*} K_0(A) \xrightarrow{\pi*} K_0(A/J) \xrightarrow{\delta} K_1(J) \xrightarrow{\iota*} K_1(A) \xrightarrow{\pi*} K_1(A/J) \xrightarrow{\delta'} K_0(J). \end{align}
What I don't get is how to write out the projection equivalent classes explicitly. As seen on page 111 of Wegge-Olsen, where the author tries to introduce $K_0$ group for $C_0(\mathbb R^2)$, unitization has to be involved. As a consequence, one get $K_0(C_0(\mathbb R^2))=\mathbb Z$ as well as $K_0(C(S^2))=\mathbb Z^2$. Then the author tries to 'explain' why there's this difference:
The only projection in $C_0(\mathbb R^2)$ is $0$, which is trivial even after stabilization. However, different story for $C(S^2)$, the projection is $0$ or $1$. At first sight, this looks like the same thing for $\mathbb C$ and one would naively imagine, when one looks into the matrix algebra $\mathbb M_\infty(C(S^2))$, one only get things like $0$, $1$, $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, etc. Therefore, naively one would conclude that $K_0$ group for $C(S^2)$ is the same as $K_0(\mathbb C)$. However, due to the exponential map, one gets 'non-trivial' maps, which contribute another $\mathbb Z$ for $K_0(C(S^2))$.
(Sorry for the long background for such a simple question.) The question is, how can we write out the projection class $[p]$, that is responsible for the extra $\mathbb Z$?
I searched a bit but all I found was proofs that $K_0(C(S^2))= \mathbb Z ^2$, which I do not doubt at all..
The exact sequence for $K_0$ splits (because it ends in $\mathbb{Z}$), which means the projection you're looking for is the one that generates $K_0(C_0(\mathbb{R}^2))$. This is sometimes called the Bott projection and you can find it here, it is $b(z)-b(\infty)$.