[proofplan] We first forget the holomorphic structure and split the short exact sequence as a sequence of smooth complex vector bundles by choosing a Hermitian metric on $E$. This identifies $E$ smoothly with $E' \oplus E''$. We then choose Chern connections on $E'$ and $E''$, form their direct-sum connection, and transfer it to $E$ through the smooth splitting. Chern-Weil invariance says that the total Chern class is independent of the chosen connection, while the curvature of the direct-sum connection is block diagonal, so the determinant polynomial factors into the product of the two determinant polynomials. [/proofplan]

[step:Split the exact sequence as smooth complex vector bundles]Since a complex manifold is paracompact, the standard partition-of-unity construction gives a smooth Hermitian metric on every smooth complex vector bundle over $X$. Choose such a metric $h_E$ on the smooth complex vector bundle underlying $E$. The holomorphic injection $\iota: E' \to E$ is, in particular, a smooth injective bundle map onto a smooth complex subbundle $\iota(E') \subset E$. Let $S \subset E$ denote the $h_E$-orthogonal complement of $\iota(E')$. For each point $x \in X$, exactness gives \begin{align*} \ker(\pi_x)=\operatorname{im}(\iota_x). \end{align*} Since $S_x$ is a vector-space complement to $\operatorname{im}(\iota_x)$ in $E_x$, the restricted map $\pi_x|_{S_x}: S_x \to E''_x$ is injective and surjective. Hence $\pi|_S: S \to E''$ is a smooth vector bundle isomorphism. Let $\sigma: E'' \to S \subset E$ denote the inverse smooth bundle map. Define the smooth complex vector bundle map $\Phi: E' \oplus E'' \to E$ by \begin{align*} \Phi(u',u'')=\iota(u')+\sigma(u'') \end{align*} for $u' \in E'_x$ and $u'' \in E''_x$ over the same point $x \in X$. The preceding paragraph shows that $\Phi_x: E'_x \oplus E''_x \to E_x$ is a complex-linear isomorphism for every $x$, and smoothness follows from the smoothness of $\iota$ and $\sigma$. Thus $E$ is smoothly isomorphic to $E' \oplus E''$ through $\Phi$.[/step]

[guided]The holomorphic exact sequence need not split holomorphically, so we do not try to construct a holomorphic isomorphism $E \cong E' \oplus E''$. Chern-Weil theory only needs smooth connections on the underlying smooth complex vector bundles, so a smooth splitting is enough. Because $X$ is a complex manifold, it is paracompact, and the standard partition-of-unity construction gives smooth Hermitian metrics on smooth complex vector bundles over $X$. Choose a smooth Hermitian metric $h_E$ on $E$. Since $\iota: E' \to E$ is a holomorphic bundle injection, it is also a smooth bundle injection, and its image $\iota(E')$ is a smooth complex subbundle of $E$. Let $S \subset E$ be the $h_E$-orthogonal complement of $\iota(E')$. Then every fiber decomposes as a direct sum of complex vector spaces: \begin{align*} E_x=\operatorname{im}(\iota_x)\oplus S_x. \end{align*} Now use exactness. For each $x \in X$, the fiber sequence \begin{align*} 0 \longrightarrow E'_x \xrightarrow{\iota_x} E_x \xrightarrow{\pi_x} E''_x \longrightarrow 0 \end{align*} is exact, so $\ker(\pi_x)=\operatorname{im}(\iota_x)$. Therefore $\pi_x$ kills exactly the first summand in the decomposition above. Its restriction $\pi_x|_{S_x}: S_x \to E''_x$ is injective because $S_x \cap \operatorname{im}(\iota_x)=\{0\}$, and it is surjective because $\pi_x: E_x \to E''_x$ is surjective. Hence $\pi_x|_{S_x}$ is a complex-linear isomorphism for every $x$. The maps $\pi|_S: S \to E''$ and its inverse $\sigma: E'' \to S$ are smooth bundle maps. Define $\Phi: E' \oplus E'' \to E$ by \begin{align*} \Phi(u',u'')=\iota(u')+\sigma(u''). \end{align*} For each fiber this map sends the two summands isomorphically onto the two complementary summands $\operatorname{im}(\iota_x)$ and $S_x$. Hence $\Phi$ is a smooth complex vector bundle isomorphism. This is the only splitting needed in the proof.[/guided]

[step:Choose a direct-sum connection and compare Chern-Weil representatives] By the same Hermitian-metric existence theorem, choose smooth Hermitian metrics $h_{E'}$ on $E'$ and $h_{E''}$ on $E''$. Let $\nabla'$ denote the Chern connection on the holomorphic Hermitian bundle $(E',h_{E'})$, and let $\nabla''$ denote the Chern connection on the holomorphic Hermitian bundle $(E'',h_{E''})$. Thus $\nabla'$ and $\nabla''$ are smooth complex connections whose $(0,1)$ parts are the Dolbeault operators of the corresponding holomorphic bundles and which are compatible with the chosen Hermitian metrics. Define the direct-sum connection $\nabla^\oplus$ on $E' \oplus E''$ by \begin{align*} \nabla^\oplus(s',s'')=(\nabla' s',\nabla'' s'') \end{align*} for smooth sections $s' \in \Gamma(E')$ and $s'' \in \Gamma(E'')$. Transport $\nabla^\oplus$ to a smooth connection $\nabla^E$ on $E$ through the smooth bundle isomorphism $\Phi: E' \oplus E'' \to E$. Thus, for a smooth section $s \in \Gamma(E)$, \begin{align*} \nabla^E s=\Phi\bigl(\nabla^\oplus(\Phi^{-1}s)\bigr). \end{align*} Here $\Phi$ is extended in the standard way to bundle-valued forms. For any smooth connection $\nabla$ on a smooth complex vector bundle, its Chern-Weil representative is the total Chern form obtained by applying the standard determinant polynomial to the curvature of $\nabla$. By Chern-Weil invariance of Chern classes under change of connection, the de Rham class represented by the total Chern form of $\nabla^E$ is $c(E)$, and the classes represented by the total Chern forms of $\nabla'$ and $\nabla''$ are $c(E')$ and $c(E'')$, respectively. This uses the standard Chern-Weil theorem for complex vector bundles: changing the connection changes the Chern forms by exact transgression forms, so the cohomology class is independent of the chosen smooth connection. [/step]

[step:Factor the determinant polynomial for the block diagonal curvature] Let $F_{\nabla'} \in \Omega^2(X;\operatorname{End}(E'))$, $F_{\nabla''} \in \Omega^2(X;\operatorname{End}(E''))$, and $F_{\nabla^\oplus} \in \Omega^2(X;\operatorname{End}(E' \oplus E''))$ denote the curvature forms of $\nabla'$, $\nabla''$, and $\nabla^\oplus$. Because $\nabla^\oplus$ acts separately on the two summands, its curvature is block diagonal: \begin{align*} F_{\nabla^\oplus}=F_{\nabla'}\oplus F_{\nabla''}. \end{align*} The total Chern form of a connection $\nabla$ on a rank $r$ complex vector bundle is the even differential form \begin{align*} c(\nabla)=\det\left(I_r+\frac{i}{2\pi}F_\nabla\right), \end{align*} where the determinant is evaluated by the invariant determinant polynomial on matrix-valued differential forms. In a local smooth frame adapted to the decomposition $E' \oplus E''$, the endomorphism-valued form $I+\frac{i}{2\pi}F_{\nabla^\oplus}$ is block diagonal with diagonal blocks $I_{E'}+\frac{i}{2\pi}F_{\nabla'}$ and $I_{E''}+\frac{i}{2\pi}F_{\nabla''}$. Multiplicativity of the determinant on block diagonal matrices gives \begin{align*} c(\nabla^\oplus)=\det\left(I_{E'}+\frac{i}{2\pi}F_{\nabla'}\right)\det\left(I_{E''}+\frac{i}{2\pi}F_{\nabla''}\right). \end{align*} Therefore \begin{align*} c(\nabla^\oplus)=c(\nabla')c(\nabla''). \end{align*} Since $\nabla^E$ is the transport of $\nabla^\oplus$ through $\Phi$, the curvature forms are conjugate under $\Phi$, and the determinant polynomial is invariant under conjugation. Hence \begin{align*} c(\nabla^E)=c(\nabla^\oplus)=c(\nabla')c(\nabla''). \end{align*} [/step]

[step:Pass from Chern forms to de Rham cohomology classes] Taking de Rham cohomology classes in the identity \begin{align*} c(\nabla^E)=c(\nabla')c(\nabla'') \end{align*} gives \begin{align*} [c(\nabla^E)]=[c(\nabla')]\,[c(\nabla'')] \end{align*} in $H_{\mathrm{dR}}^{2\bullet}(X;\mathbb{C})$. By Chern-Weil invariance, these classes are respectively $c(E)$, $c(E')$, and $c(E'')$. Thus \begin{align*} c(E)=c(E')c(E'') \end{align*} in $H_{\mathrm{dR}}^{2\bullet}(X;\mathbb{C})$, which is the desired Whitney product formula. [/step]