On a compact Riemannian manifold $M$ the Hodge decomposition takes the form $$\Omega^k(M)=d\Omega^{k-1}(M)\oplus\mathcal{H}(M)\oplus d^*\Omega^{k+1}(M)$$ Where $d^*$ is the adjoint of $d$ w.r.t. the inner product induced by the metric.
Now on a Riemann surface we do not have a metric, but we do have a canonical conformal structure. This conformal structure is enough to give us a Hodge-$\star$ operator $$\star:\Omega^1(M)\to\Omega^1(M)$$ In this setting the Hodge decomposition takes on the form $$\Omega^1_2(M)=E\oplus\star E\oplus\mathfrak{h}(M)$$ Where $$E=\overline{\{df\mid f\in \Omega^0(M)\}}$$ $$\star E=\overline{\{\star df\mid f\in \Omega^0(M)\}}$$ $$\mathfrak{h}(M)=\{\omega\in\Omega^1(M)\mid d\star \omega=d\omega=0\}$$ Where closures are in the $L^2$ sense.
These two decomposition looks quite different to me. The term $d\Omega^{k-1}(M)$ looks like $E$. But I do not recognise an analogue of $d^*$ in the Riemann surface case. $$d^*:\Omega^k(M)\to \Omega^{k-1}(M)$$ is usually constructed as $\star d\star$, however this requires a $\star:\Omega^2(M)\to\Omega^0(M)$, but the conformal structure on $M$ is not enough to give us this $\star$.
Also, the $\mathcal{H}$ in the Riemannian manifold case is characterised by a condition involving $d^*$, which we do not have on a Riemann surface, so I don't see how $\mathcal{H}$ compares to $\mathfrak{h}$.
So my question is if these two forms of the Hodge decomposition are in some sense the same, or if these are fundamentally different decompositions? The absence of a metric on the Riemann surface suggests that we should get a slightly coarser decomposition, since a metric is more rigid then a conformal structure. However there are enough similarities between the decompositions to make me think there must be some connection.
The hodge star is an isomorphism $\star \colon \Omega^k(M) \rightarrow \Omega^{n-k}(M)$. If you apply this observation with $n = k = 2$, you will see that
$$ d^{*}(\Omega^2(M)) = \{ \star d \star \omega \, | \, \omega \in \Omega^2(M) \} = \{ \star df \, | \, f \in \Omega^{0}(M) \} = \star \{ df \, | \, f \in \Omega^{0}(M) \}. $$
If you apply it to $n = 2, k = 1$, you will see that
$$ d^{*}(\omega) = \star (d (\star \omega)) = 0 \iff d (\star \omega) = 0 $$
and so $\mathfrak{h}(M) = \mathcal{H}(M)$. Thus, you get the same direct sum decomposition.