= Solution
For a pair $(L,t)$ choose an oriented basis $(\omega_1,\omega_2)$ with $\operatorname{Im}(\omega_1/\omega_2)>0$ and $t=\omega_2/N$ modulo $L$. Such a basis exists because an exact-order point gives a primitive vector modulo $N$, which can be completed to a determinant-one basis. Define the <marked-lattice model of a modular form> by
$$
\boxed{F_f(L,t)=\omega_2^{-k}f(\omega_1/\omega_2).}
$$
Changing to a basis with the same marked point uses a <matrix> with $c\equiv0$, $d\equiv1$ and therefore $a\equiv1\pmod N$, exactly $\Gamma_1(N)$. The weight-$k$ transformation of $f$ cancels the factor from $\omega_2^{-k}$, proving independence of the basis.
The resulting function has homogeneity $F_f(uL,ut)=u^{-k}F_f(L,t)$ for $u\in\mathbb C^\times$. Conversely evaluating at $(\mathbb Z\tau+\mathbb Z,1/N)$ recovers $f$. Holomorphy in $\tau$ and holomorphy in the local parameters of degenerating lattices at every <modular cusp> characterize the functions arising from <modular forms>, rather than arbitrary homogeneous lattice functions.
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