$||a||$ denotes a specific function, called norm, defined on a vector space. It maps a $n$-dimensional element of a vector space into a non-negative real number. $||a||_p$ denotes a yet particular norm defined on the vector space. Let $V$ be a vector space. Any function $p:V\to R_+$, also denoted $p(v)\equiv ||v||$ such that
- $ p $ is finite and convex
- $ p(x)=0 \implies x=0 $
- $ \forall \alpha_{}\in R, \forall x\in V, p(\alpha_{}x)=|\alpha_{}|p(x) $
is called a norm in $V$ and $(V,p)\equiv (V,||\cdot||$ is then called a normed space. You can check that your function satisfies all these properties. In your example, also, $V$ is a space of functions, that is $a_i:T\to T'$. That is a generalization of the Euclidean space (with Euclidean norm) that you may be familiar with, which is just a particular case of normed space where the underlying set is the (n-dimensional) real numbers and the norm is the called Euclidean norm, a particular case of the function that appears in your question.
For instance, the euclidean plane is a normed space such that $V=R^2$, $x=(x_1,x_2)\in R^2$, and define the norm on $R^2$ as $p(x)=||x||_2=||x||=\sqrt{(x_1+x_2)^2}=(\sum_{i=1}^2x_i^2)^{1/2}$. So it is just a plane and the norm gives the "magnitude" of the vector. Note that it is just a special case of the norm you mentioned such that $n=2, p=2, a_i(x)=x_i$, and you don't need the absolute value operator because it is a sum of squared terms.
Those topics are covered either in Real Analysis or Linear Algebra (in a more restricted way) textbooks under the rubric of norms or normed spaces.