$\Theta$ is induced by the inclusion map $Z_p\otimes Z_q' \to (C\otimes C')_{p+q}$? Let $C$ and $C'$ be two chain complexes and $C\otimes C'$ their tensor product. Then define the map
\begin{align}
\Theta : H_p(C)\otimes H_q(C')&\to H_{p+q}(C\otimes C')\\
{[z_p]}\otimes {[z_q']}&\mapsto {[z_p\otimes z_q']} 
\end{align}
What does the following mean:

$\Theta$ is induced by the inclusion map $Z_p\otimes Z_q' \to (C\otimes
 C')_{p+q}$

And why does this imply that it is natural with respect to chain maps? What does this even mean? (Does it mean that $H_p(-)\otimes H_q(-)\implies H_{p+q}(-\otimes -)$ is a natural transformation between the functors $\textbf{Ch}\times\textbf{Ch}\to \textbf{Ab}$?)
 A: Recall the definition of the tensor product of two chain-complexes $(C, \partial_C)$ and $(D, \partial_D)$, namely $$(C \otimes D)_n = \bigoplus_{i = 0} ^n C_i \otimes D_{n-i}$$
so in particular $C_p \otimes D_q \subset (C\otimes D)_{p+q}$. The boundary map $\partial_{C \otimes D}$ is defined on an element $c_p\otimes d_q$ by 
$$\begin{align}
\partial_{C\otimes D}(c_p \otimes d_q) &= \partial_C c_p\otimes d_q+(−1)^{p}c_p \otimes \partial_D d_q \\&\in (C_{p-1} \otimes D_q)\oplus (C_p \otimes D_{q-1}) \\ &\subset (C\otimes D)_{p+q-1}
\end{align}$$
(The boundary map is then extended by linearity to all of $C\otimes D$ because it has a basis consisting of homogenous elements.)
A priori if we restrict this inclusion map to cycles then we get a map $Z(C)_p \otimes Z(D)_q \to (C\otimes D)_{p+q}$, but if $\partial_C(c_p) = 0 =  \partial_D(d_q)$ if follows from definition that $\partial_{C\otimes D}(c_p\otimes d_q) = 0$, so $Z(C)_p\otimes Z(D)_q$ is actually in $Z(C\otimes D)_{p+q}$. You should verify for yourself why it follows that $\Theta$ induces a well-defined function on homology groups.
By "natural with respect to chain maps" I think they mean the following: Suppose $\phi_C\colon C\to C'$ and $\phi_D \colon D \to D'$ are chain maps, and so $\phi_C \otimes \phi_D\colon C\otimes D \to C'\otimes D'$ is also a chain map. Then the following diagram commutes: 
$$\require{AMScd}
\begin{CD}
H_p(C)\otimes H_q(D)  @>{\Theta}>> H_{p+q}(C\otimes D)\\
@V{[\phi_C]\otimes [\phi_D]}VV @V[\phi_C \otimes \phi_D]VV \\
H_p(C')\otimes H_q(D')  @>{\Theta}>> H_{p+q}(C'\otimes D')\\
\end{CD}$$
I haven't written out the details but it looks like verifying commutativity is purely formal.
