Differential Forms on a complex manifold Given a complex manifold $M$, we can complexify its wedge product of tangent cotangent bundles. Consider $\wedge^{k}T^{\ast}M \otimes \mathbb C$, and we know that $T^{\ast}M \otimes \mathbb C$ decomposes into $T^{\ast}M^{(1,0)} \oplus T^{\ast}M^{(0,1)}$. 
Then by the commutativity of wedge product and direcsum, we conclude that $\wedge^{k}T^{\ast}M = \bigoplus_{p+q = k} \wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)}$. 
A $(p,q)$ form is  a section of $\wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)}$, but why does a typical element of (p,q) form looks like $dz_{k_1} \wedge dz_{k_p} \ldots \wedge d\bar{z_{t_1}} \ldots \wedge d \bar{z_{t_1}}$ in local coordinates? Where does the tensor product go?
 A: You wrote:$$\Lambda^{k}T^{\ast}M = \bigoplus_{p+q = k} \Lambda^{p}{(T^{1,0})}^*M \otimes \Lambda^{q}{(T^{0,1})}^*M$$
This is wrong, the correct identity is
$$(\Lambda^{k}T^{\ast}M)\otimes \mathbb{C} = \bigoplus_{p+q = k} \Lambda^{p}{(T^{1,0})}^*M \wedge \Lambda^{q}{(T^{0,1})}^*M$$
You derive this identity like so:
$$\begin{align}
(\Lambda^{k}T^*M) \otimes \mathbb{C} &= \Lambda^{k} (T^*M \otimes \mathbb{C})\\
& = \Lambda^{k} ({(T^{1,0})}^*M \oplus {(T^{0,1})}^*M)\\
&= \bigoplus_{p+q = k} \Lambda^{p}{(T^{1,0})}^*M \wedge \Lambda^{q}{(T^{0,1})}^*M
\end{align}
$$
The tensor product didn't go anywhere because it wasn't there to begin with: it was already gone as soon as you wrote $T^*M \otimes \mathbb{C} = {(T^{1,0})}^*M \oplus {(T^{0,1})}^*M$.

EDIT:
As TedShifrin points out in the comment, most texts write a tensor product (as you did) instead of a wedge there. I suppose that in general $V \wedge W$ does not make sense when $V$ and $W$ are different vector spaces but $V \otimes W$ does, which probably explains the choice of "most texts". But they're wrong: $dz \otimes d\bar{z}$ is not the same as $dz \wedge d\bar{z}$. The point is that $dz \wedge d\bar{z}$ does make sense, because $dz$ and $d\bar{z}$ are both one-forms (complexified). More generally, $\Lambda^{p}{(T^{1,0})}^*M$ and $\Lambda^{q}{(T^{0,1})}^*M$ are not the same vector space but they are both subspaces of the complexified exterior algebra $(\Lambda T^*M) \otimes \mathbb{C}$, so their wedge product is a well-defined subspace of $(\Lambda T^*M) \otimes \mathbb{C}$.
A: From the identity $T^{\ast}M\otimes \mathbb{C} = T^{\ast}M^{(1,0)} \oplus T^{\ast}M^{(0,1)}$ and with fixed local frames $\{dz_1 , \dots, dz_n \}$ and $\{d\bar z_1 , \dots, d\bar z_n \}$ for $T^{\ast}M^{(1,0)}$ and $T^{\ast}M^{(0,1)}$ respectively, we can construct the bundle
$$
\bigwedge^\ast (T^{\ast}M\otimes \mathbb{C})
$$
wich is has a frame given by the wedge products of $\{dz_1 , \dots, dz_n, d\bar z_1 , \dots, d\bar z_n \}$. 
Here no tensor product will apear when you pick an element of type $(p,q)$.
The formula for the exterior power of a direct sum
$$
\wedge^{k}(T^{\ast}M\otimes \mathbb{C}) \, {\Huge \simeq} \bigoplus_{p+q = k} \wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)}
$$
is correct. The missing thing is an identification. Here a $(p,q)$ form is a section 
$$
\eta =\sum\limits_{|I|=p, |J|=q} a_{IJ}dz_I\otimes d\bar z_{J} \in \Gamma( \wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)})
$$
The identification is
$$ 
\eta \longmapsto \sum\limits_{|I|=p, |J|=q} a_{IJ}dz_I \wedge d\bar z_{J} \in \Gamma( \wedge^{p+q}(T^{\ast}M\otimes \mathbb{C})).
$$
Note that this is well defined and bijective (onto its image, $\wedge^{p,q}(T^{\ast}M\otimes \mathbb{C})$)  as you can arrange the terms so that the $dz_j$ come ahead of the $d\bar z_i$.
