The author is not claiming equality between a $d_n$ and a sum of $d_{\alpha \beta}$s. These are different types of objects, as you've noted. Rather, what is true is that the result of $d_n$ applied to the $n$-cell $e^n_\alpha$ (viewed as a generator of $H_n(X^n,X^{n-1})$) can be computed as the sum of all the $(n-1)$-cells $e^{n-1}_\beta$ (viewed as generators of $H_n(X^{n-1},X^{n-2})$), each appearing exactly $d_{\alpha \beta}$ times.
To be very explicit (and this is just linear algebra): one of the first things that we learn about the groups $C_n:=H_n(X^n,X^{n-1})$ defining the cellular chain complex is that they are isomorphic to the free Abelian groups $\mathbb{Z}\langle e^n_\alpha \rangle$ generated by the $n$-cells of the complex. Call this isomorphism $$\phi:C_n \xrightarrow{\sim} \mathbb{Z}\langle e^n_\alpha \rangle.$$
Then, as we often do in mathematics, we will treat the domain and the codomain of $\phi$ as the same object.
Under this correspondence, the data of a homomorphism $f:C_n \to A$ to another Abelian group $A$ is equivalent to the data of a homomorphism $g:\mathbb{Z}\langle e^n_\alpha \rangle \to A$ by the relation $f = g\circ \phi$. But $\mathbb{Z}\langle e^n_\alpha \rangle$ is free Abelian with basis the $n$-cells of the complex, so any homomorphism $g:\mathbb{Z}\langle e^n_\alpha \rangle \to A$ is entirely determined by its action on each basis element (each $n$-cell). This is because if you know $g(e^n_\alpha)$ for every $\alpha$, you know everything:$$g(\sum_\alpha c_\alpha e^n_\alpha) = \sum_\alpha c_\alpha g(e^n_\alpha),$$
where each $c_\alpha \in\mathbb{Z}$ is a scalar.
Similarly, the data of a homomorphism $f:C_n \to C_{n-1}$ is equivalent to the data of a homomorphism $g:\mathbb{Z}\langle e^n_\alpha \rangle \to \mathbb{Z}\langle e^{n-1}_\beta \rangle$ by the conjugation relation $f = \phi^{-1} \circ g\circ \phi$.
So, formally, the claim is that $d_n = \phi^{-1} \circ g \circ \phi$, where $g$ is defined on the basis elements by $$g(e^n_\alpha) = \sum_{\beta}d_{\alpha \beta}e^{n-1}_\beta.$$