The definition of an operad requires that that the associativity condition holds: $$f \circ (g_1(x_1^1, \dotsc, x_{n_1}^1), \dotsc, g_r(x_1^r, \dotsc, x_{n_r}^r)) = (f \circ (g_1, \dotsc, g_r)) (x_1,\dotsc, x_{n_1+\dotsb +n_r}),$$ where $f$ and $g_i$ are elements of the operad, and the $x_j$ are arbitrary inputs.
How does this definition make sense with the associative operad? If $g_1 \in \operatorname{Ass}(n)$, what is $g_1(x_1,\dotsc,x_n)$ supposed to mean?
I know that I should look at the elements of $\operatorname{Ass}$ as abstract operations. But when, for example, I want to show that $\operatorname{Ass}$ is an operad, I need to show it satisfies the associativity condition, right? How do you do that?
What you are looking at is rather the definition of an algebra over an operad. In the definition of an operad, you consider purely the operations by themselves, and not their application to "elements".
Let's work in the non-symmetric context. The operad $Ass$ has dimension $1$ in every arity, with $Ass(n)$ spanned by the operation $m_n$ representing "$(x_1,\ldots,x_n)\mapsto x_1\cdots x_n$" in an associative algebra.
Thus, what you are writing in the OP is about what happens in an algebra over the operad $Ass$.
That being said, the notation you are using is rather sloppy. It is usually better to err a bit on the formal side when working with operads since otherwise it can become quite confusing pretty quickly.