I'm missing something obvious, I suspect, but for the life of me I can't figure it out. What is wrong with the following exposition?
For any integral matrix $α$ with positive determinant let $[α]_k$ be the linear operator on the space of holomorphic functions on the extended upper-half plane $H^* = H ∪ \{i∞\}$ defined by $$f[α]_k(τ) = (\det α)^{k/2} j_α(τ) f(ατ)$$ where $j_α$ is the typical $k^{\text{th}}$ factor of automorphy of $α$ and $α$ acts on $H^*$ by usual fractional linear transformation. Then so long as $f$ is a modular form (holomorphic on $H^*$ plus $f[α]_k = f$ for all $α$ with determinant $1$) then we see $f[α]_k = f$ for all $α$ with positive determinant.
Now, consider the Hecke operator $T(n)$ of order $n$, which acts on the free abelian group of lattices via $$T(n)L = \sum_{[L:L'] = n} L'$$ Using our knowledge of bases of free abelian groups of finite rank, this equates to $$\sum αL$$ where $α$ runs over a set of representatives of the set of matrices of determinant $n$ under the action of left multiplication by $\text{SL}_2(ℤ)$ (which we know there to be $σ(n)$ of, when $σ(n)$ is the sum of the divisors of $n$). But if $F$ is the corresponding function on lattices obtained from $f$, then $$F(α[1,τ]) = F([a+bτ, c+dτ]) = F((a+bτ)[1,\dfrac{c+dτ}{a+bτ}]) =$$$$(a+bτ)^{-k}F([1,\dfrac{c+dτ}{a+bτ}]) = n^{-k/2} f[\rho α \rho^{-1}]_k(τ)$$ where $\rho = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
As such, if we convert the action of $T(n)$ to one operating on the space of modular forms of weight $k$ I should expect to calculate (where $τ∈H^*$ and $L = [1,τ]$) $$T(n)f(τ) = F(T(n)L) = F(\sum αL) = \sum F(αL) =$$$$\sum n^{-k/2} f[\rho α \rho^{-1}]_k(τ) = \sum n^{-k/2} f(τ) = n^{-k/2} σ(n) f(τ)$$
Yet this is obviously false, because $f$ was arbitrary yet I know not every modular form is a Hecke eigenform, and also that the eigenvalue $n^{-k/2} σ(n)$ is incorrect even when $f$ is an eigenform.