If these were bosonic operators $b_{\alpha}^{\dagger}$, you could rightly regard the product $b_{\alpha}^{\dagger} b_{\beta}^{\dagger}$ as a tensor product. This is because you can think of the bosonic many-body Hilbert space as the tensor product space of many harmonic oscillators, one for each $\alpha$. As a result, you can rightfully think of these operators as being defined by
$$
b_{\alpha}^{\dagger} = \mathbb{1} \otimes \ldots \otimes \mathbb{1} \otimes b^{\dagger} \otimes \mathbb{1} \otimes \ldots \otimes \mathbb{1}
$$
where the $b^{\dagger}$ appears in the $\alpha$th place; ie, it acts on the $\alpha$th harmonic oscillator mode. As a result, $b_{\alpha}^{\dagger} b_{\beta}^{\dagger}$ for $\alpha \neq \beta$ can be reasonably regarded as either an ordinary product or a tensor product, depending on how you interpret the notation.
For fermions on the other hand, you know that we cannot have the same simple tensor product structure because $c^{\dagger}_{\alpha}$ and $c^{\dagger}_{\beta}$ for $\alpha \neq \beta$ do not commute! Instead, there is a tricky phase factor which makes the operators not quite tensor products. One way to realize the fermion operators explicitly is to start with spin-half operators (ie, Pauli matrices) $\sigma^z$ and $\sigma^{\pm} = \frac{1}{2}(\sigma^x \pm i \sigma^y)$, and to write $c_{\alpha}^{\dagger}$ as
$$
c_{\alpha}^{\dagger} = \sigma^z \otimes \ldots \otimes \sigma^z \otimes \sigma^+ \otimes \mathbb{1} \otimes \ldots \otimes \mathbb{1}
$$
The string of $\sigma^z$'s on the left is often called the Jordan-Wigner string, and it implements the anticommuting algebra of the fermions. The moral of the story is that when you see an operator $c^{\dagger}_{\alpha} c^{\dagger}_{\beta}$ for $\alpha \neq \beta$, it is not quite correct to regard the product as a tensor product.
