I know that a dot product $p\cdot q$ measures the projection of parts $p$ on $q$, so that if $p$ and $q$ are orthogonal, $p\cdot q$ = 0. However, I have been wondering about whether this applies to the dot product of any arbitrary vector valued function ( with vectors as inputs) of $p$ and $q.$ Would, in general, $f(p)\cdot f(q)$ be 0 when $p$ and $q$ are orthogonal? (where $f(p)$ and $f(q)$ are any arbitrary vector valued function with vectors as inputs evaluated at $p$ and $q$)
Dot product of vector valued function
vectors
Related Solutions
Write $a \cdot b = ( |a|) ( |b| \cos \theta)$
Then consider 
The projection of B onto A is of length $|b| \cos \theta$, it is the portion of the green line up to the dotted line perpendicular to it (or I suppose beyond it). Then the dot product is the just magnitude of these two vectors, $a$ and $b\cos \theta$, multiplied.
Note: I have interchanged $a$ and $A$, $b$ and $B$ due to the diagram and your question.
Have a look in the preview of this book, Section 1.2 has a good explanation of the dot product:
If you compare your formula for $T_{ij}a_j$ to that of matrix multiplication of a matrix $A$ with $ij$ entry $A_{ij}$ and a vector $v=(v_1,...,v_n)$, you will see that $T\cdot a$ is just matrix multiplication of the matrix $T$ with entries $T_{ij}$ by the vector $a=(a_1,...,a_n)$. So the short answer to your question is 'it represents everything matrix multiplication represents'. To give a longer answer we can try to explain why we get this formula for $T a$.
Given a vector space $V$, an "$n$-tensor" refers to an element of the vector space $\otimes_{i=1}^n V$, where $\otimes$ is the tensor product. Then a vector in $v$ is a 1-tensor, and what you've called $T$ is a 2-tensor.
The dot product allows us to identify an element of $V$ with a linear map from $V\to \mathbb{R}$, because from $v\in V$ we can define the linear map $L:V\to\mathbb{R}$ by $L(w) = v\cdot w$ for any $w\in V$. This gives us an operation we can apply on pairs of vectors.
Using this fact we can identify the space of 2-tensors, $V\otimes V$ with the space of linear maps $V\to V$ by sending a pure 2-tensor $a\otimes b$ to the linear map $L_{ab}$ taking $v\to(v\cdot b) a$ and extending linearly. We get matrices from linear maps by writing the map in a basis, so if we pick a basis $\{e_i\}_{i=1}^{\dim V}$ for $V$, we get a basis of the pairs $\{e_i\otimes e_j\}$ for $V$. Then as we learned in our lin alg classes, the $ij$th entry of the matrix representing $T:V\to V$ in the $e_i$ basis is $T_{ij} = e_i \cdot (Te_j).$
So in the end we see that the dot product identifies the tensor $T$ with a linear map $V\to V$, and in a basis we recover the formula for matrix multiplication.
Best Answer
No. For example, if $p,\,q$ are $n$-dimensional vectors and $A,\,B$ are $n\times n$ matrices then $Ap\cdot Bq=p\cdot(A^TB)q$ depends on $A^TB$. In particular, we could choose $A=I$, then choose $B$ so $Bq\parallel p$, so $Ap\cdot Bq\ne0$ (unless $Bq=0$).