= Solution
First establish <rational conjugation of finite-index modular subgroups> without assuming that $\Gamma$ is a <congruence subgroup>. Multiply $\gamma$ by a positive integer to obtain an integral <matrix> $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in\Gamma(D)$, then
$$
AhA^{-1}=I+AB\operatorname{adj}(A)\in SL_2(\mathbb Z).
$$
Thus the <principal congruence subgroup> $\Gamma(D)$ is contained in $H=SL_2(\mathbb Z)\cap\gamma^{-1}SL_2(\mathbb Z)\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[SL_2(\mathbb Z):\Gamma]$. Consequently
$$
\boxed{[SL_2(\mathbb Z):\Gamma']<\infty,
\quad\Gamma'=SL_2(\mathbb Z)\cap\gamma^{-1}\Gamma\gamma.}
$$
This argument does not assert that an arbitrary <finite-index subgroup> contains a <principal congruence subgroup>.
Use the <determinant-normalized slash operator>
$$
(f|_k\gamma)(z)=(\det\gamma)^{k/2}(cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right),
\quad\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix},\quad\det\gamma>0.
$$
The positive real power of the <determinant> is used; on $SL_2(\mathbb Z)$ this reduces to the usual <slash operator for modular forms>. The <automorphy factor> identity gives the right-action rule $(f|_k\gamma)|_k\eta=f|_k(\gamma\eta)$.
A <modular form on a finite-index subgroup> of integer weight $k$ is a <holomorphic function> on the <complex upper half-plane>, invariant under this weight-$k$ action of $\Gamma$, and <holomorphic at a cusp> at each of its cusps. A <cusp of a modular group> is a $\Gamma$ orbit in $\mathbb P^1(\mathbb Q)$. If $\sigma\in SL_2(\mathbb Z)$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h\sigma^{-1}\in\Gamma$. Such $h$ exists by finite index. Then $f|_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a <cusp form> means zero constant term. Using an actual translation period avoids possible signs if a smaller <width of a cusp> is defined only modulo the center, particularly in odd weights.
For <cusp holomorphy under rational slash operators>, choose $\sigma\in SL_2(\mathbb Z)$ with $\sigma\infty=\gamma\infty$, possible by completing a primitive integer pair to a determinant-one <matrix>. Then $\sigma^{-1}\gamma=\begin{pmatrix}a&b\\0&d\end{pmatrix}$, with $a/d>0$. Up to a nonzero constant factor,
$$
f|_k\gamma(z)=(f|_k\sigma)((a/d)z+b/d).
$$
The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f|_k\sigma$. It tends to zero if $f$ is a <cusp form>. Moreover $f|_k\gamma$ is invariant under $\Gamma'$: for $\eta\in\Gamma'$, $\gamma\eta\gamma^{-1}\in\Gamma$ and the right-action rule applies. Finite index gives a translation period for $\Gamma'$, so boundedness is a <removable singularity> at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational <matrix> $\gamma\rho$, with $\rho\in SL_2(\mathbb Z)$. Thus all cusp conditions hold, and
$$
\boxed{f|_k\gamma\in M_k(\Gamma').}
$$
For the <character twist by rational translations of a cusp form>, put $B_j=\begin{pmatrix}1&j/N\\0&1\end{pmatrix}$ and $K=\Gamma_1(N)\cap\Gamma_0(N^2)$. For $\eta=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in K$, direct conjugation gives
$$
B_j\eta B_j^{-1}=\begin{pmatrix}
a+jc/N&b+j(d-a)/N-j^2c/N^2\\
c&d-jc/N
\end{pmatrix}\in SL_2(\mathbb Z).
$$
Indeed $N^2\mid c$ and $a\equiv d\equiv1\pmod N$. Every $f|_kB_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a <cusp form>, for every <Dirichlet character>:
$$
\boxed{f_\chi\in S_k(\Gamma_1(N)\cap\Gamma_0(N^2)).}
$$
There is, however, \b[a missing primitivity hypothesis in the printed final expansion claim]. The exact <Fourier expansion of a modular form> is always
$$
\boxed{f_\chi(z)=\sum_{n\geq1}a_n(f)S_{\overline\chi}(n)q^n,
\quad S_{\overline\chi}(n)=\sum_{j\in(\mathbb Z/N\mathbb Z)^\times}\overline{\chi(j)}e^{2\pi inj/N}.}
$$
Values of a <Dirichlet character> on units have modulus one, so $\chi(j)^{-1}=\overline{\chi(j)}$. For unit $n$, substitution gives $S_{\overline\chi}(n)=\chi(n)g(\overline\chi)$, where $g$ is the <Gauss sum of a Dirichlet character>. For nonunit $n$, this vanishing formula requires a <primitive Dirichlet character>.
Here is its proof in that case. Choose a <prime> $p\mid(n,N)$. Primitivity supplies a unit $u\equiv1\pmod{N/p}$ with $\chi(u)\ne1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the <Chinese remainder theorem>. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline\chi}(n)=0$. The <finite Fourier transform of a primitive Dirichlet character> now gives the corrected formula
$$
\boxed{f_\chi=g(\overline\chi)\sum_{n\geq1}\chi(n)a_n(f)q^n
\quad\text{if }\chi\text{ is primitive modulo }N.}
$$
The constant is nonzero: finite exponential orthogonality gives $\sum_{n\bmod N}|S_{\overline\chi}(n)|^2=N\varphi(N)$, whereas the proved formula makes this $\varphi(N)|g(\overline\chi)|^2$. Thus $|g(\overline\chi)|=\sqrt N$.
For a concrete counterexample to the printed unrestricted claim, take $N=2$, the <principal Dirichlet character>, and $f=\Delta$. The translation sum is $\Delta(z+1/2)=\sum_n(-1)^n\tau(n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.
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