It seems that you started off well enough. Maybe for clarity, note that we may get rid of $\sigma$, assume that $x = \infty$ and what one really wants to show is that for all characters $\chi$ of finite order, which acts trivially on $\Gamma_{\infty}$ we have
$$
I = \int_{z_0}^{z_0+h} \sum_{1 \ne \gamma \in \Gamma_{\infty} \backslash \Gamma} \chi(\gamma) j(\gamma, z)^{-k} dz = 0
$$
One could do that directly (see below), but in fact Miyake already does that in the course of proving Theorem 2.6.7. I would first like to draw your attention to condition (v), right before the theorem. Note that for $\phi = 1$, if
$\sigma^{-1} = \left( \begin{array}[cc] aa & b \\ c & d \end{array} \right)$, then we get different bounds when $c = 0$ and when $c \ne 0$.
Namely, for $c = 0$, we have $\varepsilon = 0$, but for $c \ne 0$, we have the much stronger $\varepsilon = k$.
Now, consider again the functions $\phi_{\alpha}$ in the proof of Theorem 2.6.7.
In this case, $\alpha$ runs over a set of representatives for
$\Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}$ .
Note that the lower left entry of a matrix (the one we call $c$) is the same for all elements of the double coset. Moreover, an element with $c = 0$ must be in $\Gamma_{\infty}$. Therefore, for any nontrivial $\alpha$, we have $c \ne 0$, and the stronger bound, implying that for any element $\alpha \beta$ in the double coset, this is the case, and so by the same proof as in that of Theorem 2.6.7, we see that $\phi_{\alpha}$ vanishes at $\infty$ for all $\alpha \ne \Gamma_{\infty}$.
Since $F(z) = \sum_{\alpha} \phi_{\alpha} (z)$ (see 2.6.6), we see that at $\infty$ the value of $F$ coincides with that of $\phi_1 = 1$.
(*) If you would really like to evaluate the integral, here is one way to proceed:
\begin{align*}
I &= \int_{\Gamma_{\infty} \backslash \mathbb{R}} \sum_{1 \ne \alpha \in \Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}}
\sum_{\beta \in \Gamma_{\infty} \alpha \backslash \Gamma} \chi(\alpha \beta) j(\alpha \beta, z)^{-k} dz \\
&= \sum_{1 \ne \alpha \in \Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}}
\chi(\alpha) \int_{\Gamma_{\infty} \backslash \mathbb{R}} \sum_{\beta \in (\alpha^{-1} \Gamma_{\infty} \alpha \cap \Gamma_{\infty}) \backslash \Gamma_{\infty}} j(\alpha, \beta z)^{-k} dz \\
&= \sum_{1 \ne \alpha \in \Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}}
\chi(\alpha) \int_{(\alpha^{-1} \Gamma_{\infty} \alpha \cap \Gamma_{\infty}) \backslash \mathbb{R}} j(\alpha, z)^{-k} dz = 0.
\end{align*}
Here we have used that $\chi(\beta) = 1$ for $\beta \in \Gamma_{\infty}$, that $j(\alpha \beta, z) = j(\alpha, \beta z) j(\beta, z)$, that $j(\beta, z) = 1$ for $\beta \in \Gamma_{\infty}$, that the integral $\int j(\alpha,z)^{-k} dz$ converges for $\alpha \notin \Gamma_{\infty}$, and that the sum $\sum \chi(\alpha)$ vanishes. This only works when $\chi$ is non-trivial due to convergence issues, but it gives a rough idea of what one should do.
Best Answer
The usefulness of $P_m^k$ is its behavior with respect to the Petersson inner product, for $f\in S_k(\Gamma)$ $$\langle f,P_m^k \rangle =\int_{\Gamma \setminus \Bbb{H}} f(z)\overline{P_m^k(z)}y^{k-2}dxdy=\int_0^\infty \int_0^1 f(z) \exp(2i \pi m \overline{z})y^{k-2}dxdy$$ $$= \int_0^\infty e^{-4\pi m y} a_m(f) y^{k-2}dy =a_m(f) (4\pi m)^{1-k}\Gamma(k-1)$$ thus with $(f_j)$ the basis of eigenforms of $S_k(\Gamma)$ (because the Hecke operators are normal the $f_j$ are orthogonal for the Petersson inner product) $$P_m^k(z) = \sum_j \frac{\langle f,P_m^k\rangle}{\langle f,f\rangle}f_j(z)=\sum_j \frac{a_m(f_j) (4\pi m)^{1-k}\Gamma(k-1)}{\langle f_j,f_j\rangle} f_j(z)$$
Their coefficients satisfy $$a_n(f_j)a_m(f_j) = m^{1-k}\sum_{d|(m,n)}d^{k-1}a_{\frac{nm}{d^2}}(f_j)$$ summing the latter over $j$ you get your result which is in condensed form $$T_n P_m^k =m^{1-k}\sum_{d|(m,n)}d^{k-1} P_{nm/d^2}^k$$