The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with

$$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$

a short exact sequence

$$ H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0 $$

and the finite-dimensionality of $H^1 (X;\mathcal{O})$ implies the result.

Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$ (this is not easy). Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Since there is a line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (this happens if the degree of $L$ is a sufficiently large positive number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle.

The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with

$$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$

a short exact sequence

$$ H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0 $$

and the finite-dimensionality of $H^1 (X;\mathcal{O}(L))$ implies the result.

Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$. Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Since there is a line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (this happens if the degree of $L$ is a sufficiently large positive number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle.

The proof that $H^1 (X;\mathcal{O})$ is finite-dimensional (which requires quite a bit of analysis) generalizes to general line bundles. Or one can use the result for trivial line bundles plus the existence of meromorphic sections, see the answer by Francesco.

The existence of meromorphic sections is not easy to establish; it follows from Riemann-Roch. There is also a more direct argument, using the analysis involved in the proof of Riemann-Roch.

The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with

$$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$

a short exact sequence

$$ H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0 $$

and the finite-dimensionality of $H^1 (X;\mathcal{O})$ implies the result.

Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$ (this is not easy). Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Take any line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (for example, the degree of $L$ is a sufficintly large negative number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle.

The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with

$$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$

a short exact sequence

$$ H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0 $$

and the finite-dimensionality of $H^1 (X;\mathcal{O})$ implies the result.

Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$ (this is not easy). Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Since there is a line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (this happens if the degree of $L$ is a sufficiently large positive number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle.