Your first question can be answered by using the splitting principle.

If $V \to X$ is a complex vector bundle of rank two, then $c_1(S^3V) = 6c_1(V)$ and $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$.

*Proof:* By the splitting principle, there is a map $p : Y \to X$ such that $p^*$ is injective on integral cohomology and $p^*V \cong L_1\oplus L_2$, so $p^*(S^3V) \cong S^3(p^*V) \cong S^3(L_1\oplus L_2)$. In general, we have $S^n(E_1\oplus E_2) \cong \bigoplus_{i+j=n} S^i(E_1)\otimes S^j(E_2)$, so

\begin{align*}
&\, S^3(L_1\oplus L_2)\\
\cong&\, S^3(L_1)\otimes S^0(L_2)\oplus S^2(L_1)\otimes S^1(L_2)\oplus S^1(L_1)\otimes S^2(L_2)\oplus S^0(L_1)\otimes S^3(L_2)\\
\cong&\, L_1^3\oplus L_1^2\otimes L_2\oplus L_1\otimes L_2^2\oplus L_2^3.
\end{align*}

It follows that $c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1) + 6c_1(L_2) = 6c_1(L_1\oplus L_2)$. So $$p^*c_1(S^3V) = c_1(p^*S^3V) = c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1\oplus L_2) = 6c_1(p^*V) = p^*(6c_1(V)).$$ By the injectivity of $p^*$, we have $c_1(S^3V) = 6c_1(V)$.

Similarly, one can compute that

\begin{align*}
c_2(S^3(L_1\oplus L_2)) &= 11c_1(L_1)^2 + 11c_1(L_2)^2 + 32c_1(L_1)c_1(L_2)\\
&= 11(c_1(L_1)+c_1(L_2))^2 + 10c_1(L_1)c_1(L_2)\\
&= 11c_1(L_1\oplus L_2)^2 + 10c_2(L_1\oplus L_2)
\end{align*}

and hence $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$. $\quad\square$

In this case, $\Sigma^-$ is an $SU(2)$ bundle and hence $c_1(\Sigma^-) = 0$. So we see that $c_1(S^3\Sigma^-) = 0$ and $c_2(S^3\Sigma^-) = 10c_2(\Sigma^-)$. Therefore

$$\operatorname{ch}(S^3\Sigma^-) = \operatorname{rank}(S^3\Sigma^-) + c_1(S^3\Sigma^-) + \frac{1}{2}(c_1(S^3\Sigma^-)^2 - 2c_2(S^3\Sigma^-)) = 4 - 10c_2(\Sigma^-).$$

To find $c_2(\Sigma^-)$, or $c_2(\Sigma^+)$, we can proceed as follows.

As $\Sigma^{\pm}$ is an $SU(2)$-bundle which is a lift of the $SO(3)$-bundle $\Lambda^{\pm}$, there is a relationship between $c_2(\Sigma^{\pm})$ and $p_1(\Lambda^{\pm})$, namely $p_1(\Lambda^{\pm}) = -4c_2(\Sigma^{\pm})$; see Appendix E of *Instantons and Four-Manifolds* by Freed and Uhlenbeck, also this question. So now we just need to know $p_1(\Lambda^{\pm})$, but this is given by $\pm 2e(M) + p_1(M)$; see Chapter $6$, Proposition $5.4$ of *Metric Structures in Differential Geometry* by Walschap. Therefore

$$c_2(\Sigma^{\pm}) = \mp\frac{1}{2}e(M) - \frac{1}{4}p_1(M).$$

Note, as $M$ is assumed to be spin, its signature is a multiple of $16$ by Rohklin's Theorem, so $\frac{1}{4}p_1(M)$ is an integral class. As the signature of $M$ is even, so is the Euler characteristic, and hence $\frac{1}{2}e(M)$ is also an integral class.

Finally, as $c_2(\Sigma^-) = \frac{1}{2}e(M) - \frac{1}{4}p_1(M)$, we see that

\begin{align*}
\int_M(10c_2(\Sigma^-) - 4)\left(1 - \frac{1}{24}p_1(M)\right) &= \int_M 10c_2(\Sigma^-) + \frac{1}{6}p_1(M)\\
&= \int_M 5e(M) - \frac{5}{2}p_1(M) + \frac{1}{6}p_1(M)\\
&= \int_M 5e(M) - \frac{7}{3}p_1(M)\\
&= 5\chi(M) - 7\tau(M)
\end{align*}

as claimed.