The question is from Farkas & Kra Riemann surfaces p.89. I won't explain some terminologies that appear in the question because it will make the post ridiculously long. But if someone asks for some details or notations via comment, I'll add them.
Let $M$ be a Riemann surface and define the vector space $H^1_{hol}(M)$ of holomorphic differentials of $M$ factored by the subspace of exact holomorphic differentials (the latter are the images of holomorphic functions under the differential operator $d$).
Theorem. Let $M$ be a compact Riemann surfaces of genus $g\geq 0$ and let $P_1,\ldots,P_k$ be $k>0$ distinct points on $M$. Set $M' = M-\{P_1,\ldots,P_k\}$. Then $$\dim H^1_{hol}(M') = 2g+k-1.$$ Further, each element of $H^1_{hol}(M')$ may be represented by an abelian differential of the third kind on $M$ that is regular on $M'$ has a pole of order at most $2g$ at $P_1$ and at most a simple pole at $P_j$, $j =2,\ldots,k$.
Proof. $\dim H^1_{hol}(M')\geq 2g+k-1$: Since the operator $d$ sends regular functions to regular differentials, meromorphic functions to meromorphic differentials, and functions with essential singularities to differentials with essential singularities, the theorem will be proved if we show that the quotient of the meromorphic $1$-forms on $M$ that are regular on $M'$ by the subspace of images under $d$ of the meromorphic functions on $M$ that are regular on $M'$ is exactly $2g+k-1$.
We use induction on $k$. Assume that $k =1$. For each integer $n$ with $n\geq 2g$, there exists a meromorphic function $f$ on $M$ which is regular on $M-\{P_1\}$ and has a pole of order $n$ at $P_1$. Therefore, every meromorphic $1$-form on $M$ which is regular except possibly at $P_1$ is equivalent modulo exact forms to one with a pole of order at most $2g$. ...
I don't understand the last statement. I understand that by Weierstrass gap theorem if $n\geq 2g$, there exists a meromorphic function $f$ on $M$ which is regular on $M-\{P_1\}$ and has a pole of order $n$ at $P_1$. But how does this imply some exactness?