This explanation is similar to Peter's, but I think it is still worth mentioning.

Let $V$ be a real vector space and let $J \in \operatorname{End}(V)$ be an almost complex structure on $V$ (i.e. $J^2 = -\operatorname{id}_V$). Then $V$ has the structure of a complex vector space by defining

$$(a + bi)v := av + bJ(v).$$

Over $\mathbb{R}$, $J$ has no eigenvalues, so we extend $J$ to $V_{\mathbb{C}} := V\otimes_{\mathbb{R}}\operatorname{id}_{\mathbb{C}}$, the complexification of $V$, by

$$J\otimes_{\mathbb{R}}\operatorname{id}_{\mathbb{C}}$$

which has eigenvalues $\pm i$. Let $V^{1,0}$ and $V^{0,1}$ be the eigenspaces corresponding to $i$ and $-i$ respectively. Then $V_{\mathbb{C}} = V^{1,0}\oplus V^{0,1}$.

By considering the action of $J\otimes_{\mathbb{R}}\operatorname{id}_{\mathbb{C}}$ on elements of the form $v\otimes 1 + w\otimes i$ we find that

$$V^{1,0} = \left\{\frac{1}{2}(v\otimes 1 - J(v)\otimes i) \mid v \in V\right\}$$

and

$$V^{1,0} = \left\{\frac{1}{2}(v\otimes 1 + J(v)\otimes i) \mid v \in V\right\}.$$

The choice of factor here is seemingly arbitrary, but it has some advantages. For example, there is a natural real subspace $V_{\mathbb{R}} := \{v\otimes 1 \mid v \in V\} \subset V_{\mathbb{C}}$ which is isomorphic to $V$. For elements of this subspace, we have the simple decomposition into its $(1, 0)$ and $(0, 1)$ parts given by

$$v\otimes 1 = \frac{1}{2}(v\otimes 1 - J(v)\otimes i) + \frac{1}{2}(v\otimes 1 + J(v)\otimes i).$$

Furthermore, the following map
\begin{align*}
\phi^{1,0} : V &\to V^{1,0}\\\\
v &\mapsto \frac{1}{2}(v\otimes 1 - J(v)\otimes i)
\end{align*}
defines an isomorphism of complex vector spaces between $(V, J)$ and $(V^{1,0}, i)$.

Now let $\{w_1, \dots, w_n\}$ be a basis for $V^{1,0}$ as a complex vector space. Then $w_j = \phi^{1,0}(v_j)$ for some $v_j \in V$. Then $\{v_1, \dots, v_n\}$ is a basis for $V$ as a complex vector space and $\{v_1, \dots, v_n, J(v_1), \dots, J(v_n)\}$ is a basis for $V$ as a real vector space.

Now consider the dual space $V^{\ast}$ of $V$ as a real vector space. The almost complex structure $J$ on $V$ induces an almost complex structure $J'$ on $V^{\ast}$ given by $J'(\varphi)(v) = \varphi(Jv)$. So we can apply all of the above constructions to $V^{\ast}$.

Let $\{\psi_1, \dots, \psi_n\}$ be the dual basis to $\{w_1, \dots, w_n\}$. Then $\psi_j = \phi^{1,0}(\varphi_j)$ for some $\varphi_j \in V^*$. Then $\{\varphi_1, \dots, \varphi_n\}$ is a basis for $V^*$ as a complex vector space and $\{\varphi_1, \dots, \varphi_n, J'(\varphi_1), \dots, J'(\varphi_n)\}$ is a basis for $V^*$ as a real vector space. Here's the surprise:

$\{\varphi_1, \dots, \varphi_n, J'(\varphi_1), \dots, J'(\varphi_n)\}$ **is not** the dual basis to $\{v_1, \dots, v_n, J(v_1), \dots, J(v_n)\}$!

In fact, $J'(\varphi_j)(Jv_k) = \varphi_j(J^2v_k) = -\varphi_j(v_k) = -\delta_{jk}$, so the dual basis is actually $\{\varphi_1, \dots, \varphi_n, -J'(\varphi_1), \dots, -J'(\varphi_n)\}$.

So if $\{\varphi_1, \dots, \varphi_n, \xi_1, \dots, \xi_n\}$ is the dual basis to $\{v_1, \dots, v_n, J(v_1), \dots, J(v_n)\}$ (i.e. $\xi_j = -J'(\varphi_j)$), then a basis for $(V^*)^{1,0}$ is

$$\left\{\frac{1}{2}(\varphi_j\otimes 1 - J'(\varphi_j)\otimes i) \mid j = 1, \dots, n\right\} = \left\{\frac{1}{2}(\varphi_j\otimes 1 + \xi_j\otimes i) \mid j = 1, \dots, n\right\}.$$