f is holomorphic at the cusps if $$(c\tau+d)^kf(\tau)=\sum_{n=0}^{\infty}b_ne^{2\pi i\frac{n}{h}\frac{a\tau+b}{c\tau+d}},$$ where $a,b,c,d$ are coefficients of a congruence subgroup. Is this equivalent to saying that $$(f|_kM)(\tau)=\sum_{n=0}^{\infty}b_ne^{2\pi i\tau n/h}?$$ I think this follows if one substitutes $\tau$ by $M^{-1}\tau.$
I do not understand from which property the following definition comes $$(f|_{k}\gamma)(\tau)=c_n\sum_{n=0}^{\infty}e^{2\pi i \tau n/h} $$ for $\gamma \in SL_2(\mathbb{Z}).$
There is some considerable confusion here. We say that $f$ is holomorphic at a cusp represented by $\frac{a}{c}$ if for any $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \textrm{SL}_2(\mathbb{Z})$ such that $\gamma(\infty) = \frac{a}{c}$ the function $$(f\mid_k \gamma)(\tau) = (c\tau + d)^{-k}f(\gamma \tau)$$ is holomorphic at $\infty$. Holomorphicity at all of the cusps amounts to the statement that $(f\mid_k \gamma)(\tau)$ is holomorphic at infinity for all $\gamma \in \textrm{SL}_2(\mathbb{Z})$.
You seem to think that you can conclude the holomorphicity of $f$ at an arbitrary cusp from the holomorphicity of $f$ at $\infty$.This is true if $f$ is modular on all of $\textrm{SL}_2(\mathbb{Z})$, since in this case there is only one cusp, but for a given congruence subgroup there may be more than one cusp and holomorphicity at a given cusp doesn't imply holomorphicity at the other cusps.