Alternative construction of the first Chern class map Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of  the holomorphic cotangent bundle.
Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove.
 A: This is an extended comment to show that in the 1-dimensional case this comes down to comparing a "Hodge-theoretic inclusion" with a connecting map. 
Your version of the analytic exponential sequence implicitly rests on the map $e^{2\pi i(\cdot)}$ (giving as kernel term the constant sheaf $\mathbf{Z}$) rather than $e^{(\cdot)}$ (giving as kernel term the constant sheaf $\mathbf{Z}(1)$). The latter is more canonical insofar as it doesn't involve a choice of basis $2\pi i$ of $\mathbf{Z}(1)$, so if we work with that at the outset then in your question the division by $2\pi i$ (presumably the same choice of $\mathbf{Z}(1)$-basis as implicit in your exponential sequence) goes away, leaving us with the more "canonical" map ${\rm{d}}\log: f \mapsto {\rm{d}}f/f$.  So we'll argue in terms of this more canonical exponential sequence (and corresponding adjustment to your question).
Observe that there is an evident map from the short exact sequence
$$0 \rightarrow \mathbf{Z}(1) \rightarrow \mathscr{O}_X \stackrel{\exp}{\rightarrow} \mathscr{O}_X^{\times} \rightarrow 1$$
to
$$0 \rightarrow \mathbf{C} \rightarrow \mathscr{O}_X 
\stackrel{\rm{d}}{\rightarrow} \Omega^1 \rightarrow 1$$
(using ${\rm{d}}\log$ along the right side and the identity map on middle terms), so this yields a commutative diagram of connecting maps
$$\begin{array}[c]{ccc}
{\rm{H}}^1(X, \mathscr{O}_X^{\times}) &{\rightarrow}& {\rm{H}}^2(X, \mathbf{Z}(1)) \\
\downarrow\scriptstyle{c'_1}&&\downarrow\\
{\rm{H}}^1(X, \Omega^1_X) &{\rightarrow}& {\rm{H}}^2(X, \mathbf{C})
\end{array}$$
where the horizontal maps are connecting maps.  
Given how you defined $c_1$, your question in the case of dimension 1 thereby comes down to asking if the connecting map along the bottom coincides with the "Hodge-theoretic inclusion" (an equality in this case).  To see that such matters are not entirely trivial, note that the "Hodge-theoretic inclusion" ${\rm{H}}^0(X,\Omega^1_X) \hookrightarrow {\rm{H}}^1(X, \mathbf{C})$ is the negative of the connecting map arising from 
$$0 \rightarrow \mathbf{C} \rightarrow \mathscr{O}_X 
\stackrel{\rm{d}}{\rightarrow} \Omega^1 \rightarrow 1.$$
Maybe from degree 1 into degree 2 there's a cancellation of signs and one gets an equality rather than a sign discrepancy?  Good luck sorting it out.
A: $\def\cO{\mathcal{O}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}$Yes, this is right. This answer will prove the equality of three maps $H^1(\cO^{\ast}) \to H^2(\CC)$. 
(a) The map $$H^1(\cO^{\ast}) \overset{c_1}{\longrightarrow} H^2(\ZZ) \overset{2 \pi i}{\longrightarrow} H^2(\CC)$$ induced from the exponential sequence and the inclusion of $\ZZ$ into $\CC$.
(b) The map $$H^1(\cO^{\ast}) \overset{d \log}{\longrightarrow} H^1(W^1) \to H^1(Z^1) \cong H^2(\CC)$$ where $W^1$ is the $d$-closed $(1,0)$-forms and $Z^1$ is the $d$-closed $1$-forms. 
(c) The map $$H^1(\cO^{\ast})  \overset{d \log}{\longrightarrow} H^1(\Omega^1) \cong H^{11} \to H^2_{DR} \cong H^2(\CC)$$ where $H^{11}$ is the harmonic $(1,1)$-forms.
We will use the following notations:
$A^{pq}$ the sheaf of smooth $(p,q)$-forms.
$W^p$ the sheaf of $d$-closed $(p,0)$ forms.
$\Omega^p$ the sheaf of holomorphic $(p,0)$ forms.
$A^k$ the sheaf of $k$-forms.
$Z^k$ the sheaf of closed $k$-forms.
$H^{pq}$ are the harmonic $(p,q)$ forms.
We have a commutative diagram with exact rows:
$$\begin{matrix}
0 & \to & \ZZ & \overset{2 \pi i}{\longrightarrow} & \cO & \overset{\exp}{\longrightarrow} & \cO^{\ast} & \to & 0 \\
& & 2 \pi i \downarrow & & = & & \downarrow d \log & & \\
0 & \to & \CC & \longrightarrow & \cO & \overset{d}{\longrightarrow} & W^1 & \to & 0 \\
& & =  & & \downarrow & & \downarrow  & & \\
0 & \to & \CC & \longrightarrow & C^{\infty} & \overset{d}{\longrightarrow} & Z^1 & \to & 0 \\
\end{matrix}$$
Exactness of the first row is that nonvanishing holomorphic functions locally have holomorphic logarithms, the last row is the Poincare lemma, and the second row is the Poincare lemma plus the fact  that integrals of holomorphic functions are holomorphic.
The boundary map on sheaf cohomology is functorial in such diagrams of short exact sequences, so we have a commutative diagram:
$$\begin{matrix}
H^1(\cO^{\ast}) & \overset{c_1}{\longrightarrow} & H^2(\ZZ) \\
d \log \downarrow & & \downarrow 2 \pi i \\
H^1(W^1) & \longrightarrow & H^2(\CC) \\
\downarrow & & = \\
H^1(Z^1) & \cong & H^2(\CC) \\
\end{matrix}$$
The last row is an isomorphism by partitions of unity.
The equality (a)=(b) expresses two ways of going around this diagram.
To relate (b) and (c), we need some generalities about $H^q(W^p)$ which I wish I knew a citation for.
We first recall that the Poincare short exact sequence
$$0 \to Z^p \to A^p \overset{d}{\longrightarrow} Z^{p+1} \to 0$$
induces isomorphisms
$$H^{k}_{DR} = H^0(Z^k)/d H^0(A^{k-1}) \cong H^1(Z^{k-1}) \cong \cdots \cong H^k(Z^0) = H^k(\CC) .$$
Key compatability: The map $W^p \to Z^p$ induces an inclusion $H^q(W^p) \to H^q(Z^p) \cong H^{p+q}(\CC)$ with image $\bigoplus_{k \leq q} H^{(p+q-k)k}$. The map $H^q(W^p) \to H^q(\Omega^p)$ induces the projection $\bigoplus_{k \leq q} H^{(p+q-k)k} \to H^{pq} \cong H^q(\Omega^p)$.
Using this, we prove (b)=(c). The image of $H^1(W^1)$ in $H^1(Z^1)$ is $H^{20} \oplus H^{11}$, so every class in $H^1(\cO^{\ast})$ is sent by (b) to a class of the form $\alpha \oplus \beta$. As you say, we can use the description from (a) to show that this class is realizable by a $(1,1)$-form, so $\alpha=0$. Then $\beta \in H^{11}$ is the image of the composition $H^1(W^1) \to H^2(\CC) \to H^{11}$, and the key compatability says that this is the same as the map $H^1(W^1) \to H^1(\Omega^1)$. So we can compute $\beta$ using the map $H^1(W^1) \to H^1(\Omega^1)$, and that is your option (c).
I'm going to hold off on writing a proof of the Key Compatability on the assumption that someone will give me a citation for it, ideally in the same sort of classical language that the OP uses in the question.
