Let $A$ and $B$ be chain complexes. Let $I$ be the chain complex with $I_0=\mathbb{Z}[a,b]$ (the $\mathbb{Z}$ linearization of the set ${a,b}$), $I_1=\mathbb{Z}[e]$ with all other $I_n$ trivial, and $\partial_1(k\cdot e)=k\cdot a+k\cdot b$.
Let $H: I \otimes A\rightarrow B$ be a chain map, I want to show that that $H^a:A\rightarrow B, H^a(x)=H(a\otimes x)$ and $H^b:A\rightarrow B, H^b(x)=H(b\otimes x)$ are also chain maps.
Attempt: To show $H^a$ is a chain map we need to show that $\partial^B_n \circ H^a_n=H^a_{n-1}\circ \partial^C_n$. Since $H$ is a chain map we know $\partial^B_n \circ H_n=H_{n-1}\circ \partial^{I\otimes C}_n$. But I'm not sure how to continue
Recall that $(I\otimes A)_n = \bigoplus_{p+q= n} I_p \otimes A_q$, so if we define $x\mapsto a\otimes x, A\to I\otimes A$, this is a well-defined map, so we only need to check that it commutes with the differential.
Now you have to also remember what the definition of $\partial^{I\otimes A}$ is : $\partial^{I\otimes A}(s\otimes x ) = \partial^I(s) \otimes x + (-1)^{|s|}s\otimes \partial^A(x)$
But here, $\partial^I(a) = 0$ and $|a| = 0$, so $\partial^{I\otimes A}(a\otimes x) = a\otimes \partial^A(x)$
If you call $f: x\mapsto a\otimes x$, this reads $\partial^{I\otimes A}\circ f (x) = f\circ \partial^A(x)$, in other words $\partial^{I\otimes A}\circ f = f\circ \partial^A$ : $f$ is a chain map.
The answer to your question follows because chain maps compose to give chain maps