Let me start with the second question:

2) The map $\alpha$ will indeed respect the degrees. You should think of it as a map between two $E_2$-algebras in unbounded chain-complexes (whose domain happens to be concentrated in non-negative degrees, while its codomain is concentrated in non-positive degrees).

1) As a result of (2), the data of the map $\alpha$ is actually equivalent to the data of a map of discrete commutative algebras $A \to B$. To see this, observe that the inclusion $\iota_{\leq 0}: {\rm Ch}_{\leq 0}(\mathbb{k}) \to {\rm Ch}(\mathbb{k})$ of non-positively graded chain-complexes into unbounded chain-complexes has a left adjoint $\tau_{\leq 0}: {\rm Ch}(\mathbb{k}) \to {\rm Ch}(\mathbb{k})_{\leq 0}$, the truncation functor. Both $\iota_{\leq 0}$ and $\tau_{\leq 0}$ are symmetric monoidal functors and descend to an adjunction between $E_2$-algebras in ${\rm Ch}(\mathbb{k})$ and $E_2$-algebras ${\rm Ch}(\mathbb{k})_{\leq 0}$. This means that the data of an $E_2$-algebra map $\alpha: HC_*(A) \to HC^*(B)$ is equivalent to the data of an $E_2$-algebra map $A = \tau_{\leq 0}HC_*(A) \to HC^*(B)$. A similar argument using cotruncation instead of truncation shows that the data of such a map is equivalent to the data of an $E_2$-algebra map $A \to \tau^{\geq 0}HC^*(B) = B$. Since $A$ and $B$ are discrete this is the same as a map of commutative algebras.

3) Suppose $A$ was an $E_1$-algebra and $B$ a chain-complex (which we can think of as an $E_0$-algebra object in ${\rm Ch}(\mathbb{k})$). Then the structure of an $A$-module on $B$ is given by a map of $E_1$-algebras $A \to {\rm End}(B)$. This is because the monoidal structure on ${\rm Ch}(\mathbb{k})$ is closed, i.e., admits a compatible system of internal mapping objects. If $A$ is now an $E_2$-algebra and $B$ is an $E_1$-algebra, then we can think of $A$ as an $E_1$-algebra in $E_1$-algebras, and so there is an associated notion of an action of $A$ on $B$ (in $E_1$-algebras) in which case one usually says that $B$ has the structure of an $A$-algebra. However, the induced tensor product on $E_1$-algebras is not a closed monoidal structure: given two $E_1$-algebras $B,B'$, there is no internal mapping object ${\rm Map}(B,B')$ in $E_1$-algebras. Nonetheless, there is still an object which controls $A$-actions on $B$, and that is the center $HC^*(B)$ of $B$, considered as an $E_2$-algebra. In particular, $A$-algebra structures on $B$ are classified by $E_2$-algebra maps $A \to HC^*(B)$. Note that for $A$, as an $E_1$-algebra object ${\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$, there is also the associated notion of an *$A$-bimodule* structure on a given $B \in {\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$. This, in turn, can be described by a map of $E_2$-algebras $A^{\rm op} \otimes A \to HC^*(B)$. Now let's consider the case at hand where $A$ is an $E_3$-algebra and $B$ is an $E_1$-algebra. Then we can think of $A$ as an $E_2$-algebra in $E_1$-algebras. Now, just like associative algebras have the associated notion of a bimodule, for $E_2$-algebras there is the analogous notion of an *$E_2$-module*. Roughly speaking, an $E_2$-action of $A$ on $B$ means a "continuous family" of $A$-actions on $B$ parameterized by the circle (with suitable compatibilities with the $E_2$-structure of $A$). If $A$ is an $E_2$-algebra in ${\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$ and $B$ is an object of ${\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$ then the data of an $E_2$-action of $A$ on $B$ will be given by an $E_2$-algebra map of the form $\alpha: HC_*(A) \to HC^*(B)$. When $A$ and $B$ are discrete commutative algebras this is the same as just an $A$-algebra structure on $B$, which, in turn, is given by a map of commutative algebras $A \to B$.