= Solution
Write $w=2k$. For a positive-determinant real <matrix> $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, extend the <slash operator for modular forms> by
$$
(f|_wA)(\tau)=(\det A)^{w/2}(c\tau+d)^{-w}f(A\tau).
$$
Let $\mathcal M_n$ be the integral <matrices> of <determinant> $n>0$. Each left $SL_2(\mathbb Z)$ <orbit> has a unique representative $\begin{pmatrix}a&b\\0&d\end{pmatrix}$ with $ad=n$, $a,d>0$, $0\leq b<d$. To obtain it, the <Euclidean algorithm> on the first column produces $(a,0)$ with $a>0$; the <determinant> fixes $d$, and a row shear reduces $b$ modulo $d$. This also proves uniqueness.
Define the normalized <Hecke operator>
$$
\boxed{T_w(n)f=n^{w/2-1}\sum_{A\in SL_2(\mathbb Z)\backslash\mathcal M_n}f|_wA
=n^{w-1}\sum_{ad=n}d^{-w}\sum_{b=0}^{d-1}f\left(\frac{a\tau+b}{d}\right).}
$$
Right multiplication by $SL_2(\mathbb Z)$ permutes these left orbits, and the slash operation is associative. Hence the sum satisfies the same <modular form> transformation law. It is <holomorphic> on $\mathbb H$ since each argument stays there.
For $f=\sum_{r\geq0}c(r)q^r$, summing the phase $e^{2\pi irb/d}$ over $b$ gives zero unless $d\mid r$, and gives $d$ when $d\mid r$. For the resulting term to contribute to $q^m$, one needs $a\mid m$ and $r=dm/a=nm/a^2$. Its scalar factor is $n^{w-1}d^{1-w}=a^{w-1}$. Thus
$$
\boxed{T_w(n)f=\sum_{m\geq0}\left(\sum_{a\mid\gcd(n,m)}a^{w-1}c\left(\frac{nm}{a^2}\right)\right)q^m,}
$$
where $\gcd(n,0)=n$. All powers are nonnegative, so the sum is <holomorphic> at infinity and belongs to $M_w$; level one has only this <modular cusp> class. Its constant coefficient is $\sigma_{w-1}(n)c(0)$, so <modular cusp> forms are preserved. This is the full <Fourier coefficients of a composite-index Hecke operator> formula, including the constant term.
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