Since $f$ is a modular form of weight k, it satisfies $$f(M\tau)=(c\tau+d)^kf(\tau)$$, where $M\tau$ denotes the usual Mobius-transformation. Is the following approach right?
$$f(M\tau)=(c\tau+d)^kf(\tau) \Leftrightarrow \frac{1}{f(M\tau)}=(c\tau+d)^{-k}\frac{1}{f(\tau)}$$
So $\frac{1}{f}$ would satisfy the equation and further would be meromorphic, since we can take the set of roots from f away from $\mathbb{H}$?
Which is open, too, is how to justify the fourier-expansion. Is it equivalent to 1/f having a root in $\infty$?
A function on $\mathbb{H} : Im(\tau) > 0$ is said modular of weight $k$ (for the full modular group $\Gamma \simeq SL_2(\mathbb{Z})$) iff for every $\gamma \in \Gamma$ : $\forall \tau \in \mathbb{H},\ f(\tau) = f |_k\gamma(\tau) \overset{def}=f(\gamma \tau) (c\tau+d)^{-k} $.
It is said to be a meromorphic modular form iff it is meromorphic on $\mathbb{H}$ and also at the unique cusp $i\infty$ of $\mathbb{C} / \Gamma$, that is as $\tau \to i\infty$ : $f(\tau) \sim e^{2i \pi m \tau}$ for some $m \in \mathbb{Z}$. When $m \ge 0$ we call it a modular form, and when $m > 0$ a cusp form.
As you noticed, for every $\gamma \in \Gamma$ : $\forall \tau \in \mathbb{H}, \frac{1}{f(\gamma \tau)} (c\tau+d)^{k} = \frac{1}{f(\tau)}$ that is $f$ modular of weight $k$ implies $1/f$ modular of weight $-k$.
Also, $f$ meromorphic on $\mathbb{H}$ means that $1/f$ is meromorphic on $\mathbb{H}$, so all we have to check is that $1/f$ is meromorphic at the cusp.
And for this, look at $g(q) = f(\frac{\log(q)}{2i \pi})$ that is a function defined for $0 < |q| < 1$. Note how $f(\tau+1) = f(\tau)$ and meromorphic on $\mathbb{H}$ means that $g(q)$ is well-defined and meromorphic on $0 < |q| < 1$ (not depending on the branch of the logarithm you choose, see this example plot of $g(q)$ for $f$ an Eiseinstein series)
Now $f$ meromorphic at the cusp means that $g(q)q^m$ has a removable singularity at $q=0$, i.e. $g(q)$ is meromorphic around $q= 0$, and so is $\frac{1}{g(q)}$, thus it has a Laurent series $$\frac{1}{g(q)} = \sum_{n = -m}^\infty c_n q^n$$ and since $\frac{1}{f(\tau)} = \frac{1}{g(e^{2i \pi \tau})}$ you have that $\frac{1}{f(\tau)} = \sum_{n = -m}^\infty c_n e^{2i \pi n \tau}$ i.e. $1/f$ is meromorphic at the cusp.
There is not much more to say, except that the radius of convergence of $\sum_{n = -m}^\infty c_n q^n$ is the absolute value of the first zeros of $g(q)$, that in general is not $1$, for example take $f(\tau) = G_{2k}(\tau), k \ge 6$ an Eisenstein series having some zeros on $\mathbb{H}$.