Tensors Introduction For orthonormal right-handed bases {$\underline{e_i}$} and {$\underline{e'_i}$} in $\mathbb{R^3}$ with corresponding Cartesian coordinates {$x_i$} and {$x'_i$}: $\underline{x} = x_i\underline{e_i} = x'_i\underline{e'_i}$. Any such bases are related by a rotation: 
$$\underline{e'_i} = R_{ip}\underline{e_i}, \  x'_i = R_{ip}x_p$$
I am able to derive this relation, but when we go on to say that two matrices are related by 
$$A'_{ij} = R_{ip}R_{jq}A_{pq}$$
I don't see where this comes from, and also when we generalise to tensors with n indices why do we need to apply R n times? 
Also do we every have to apply different transformations to the same tensor? (as in can we have $T'_{ij} = A_{ip}B_{jq}T_{pq}$ for $A,B$ distinct?
 A: Let me try to answer your questions in two ways: one sort-of down to earth, and one that's more formal.
For a down to earth explanation, a rank 2-tensor is a bilinear map that takes in $2$ vectors. Changing basis $e_j'=R_{ij}e_i$ should be done to each slot of the bilinear map - hence two factors of $R$. 
For a higher rank tensor, you should pick up the number of $R$'s that matches the number of slots that the multilinear map has. So if you had a rank 3 tensor, you would have 3 slots, and hence a change of basis would introduce three factors of $R$.

Here's the more formal explanation.
The idea is that when you write $A_{ij}$, these numbers really are defining an object $A= \sum_{i,j} A_{ij} e_i\otimes e_j$ in the vector space $\mathbb{R}^3\otimes \mathbb{R}^3$, where $\{e_i\}$ is a basis for $\mathbb{R}^3$. We would like $A$ to be the same object no matter which basis we pick (read: independent of coordinate system), so if $e_i'$ is a new basis, we want to say that
$$
  \sum_{ij} A_{ij}' e_{i}'\otimes e_j' = A = \sum_{i,j} A_{ij} e_i \otimes e_j. \qquad (*)
$$
Inserting the relation you derived: $e_p' = R_{ip}e_i$ twice into the left-hand-side of $(*)$ we see that
$$
\begin{align*}
  \sum_{p,q} A_{pq}' e_{p}'\otimes e_q' &= \sum_{p,q} A_{pq}' \left(\sum_{i}R_{ip}e_i\right) \otimes \left(\sum_j R_{jq} e_j\right)\\
&= \sum_{i,j} \left(\sum_{p,q} R_{ip} R_{jq} A_{pq}'\right) e_i\otimes e_j
\end{align*}
$$
Therefore, again by $(*)$, you have an equality
$$
 \sum_{i,j} \left(\sum_{p,q} R_{ip} R_{jq} A_{pq}'\right) e_i\otimes e_j=\sum_{i,j} A_{ij} e_i \otimes e_j
$$
and since $\{e_i\otimes e_j\}$ is a basis, we see that
$$
 \left(\sum_{p,q} R_{ip} R_{jq} A_{pq}'\right) =A_{ij}. \qquad (**)
$$
This explains why when you have a rank-two tensor you need two factors of $R$. This reasoning also extends two higher-rank tensors: if $A$ were instead a rank-three tensor, for example, you would have an expression
$$
  A = \sum_{i,j,k} A_{ijk} e_i\otimes e_j \otimes e_k
$$
and the same computation as above would introduce three factors of $R$.
Hopefully this process also demonstrates for you that your second question is a little odd. We derived the transformation rule $(**)$ by inserting the same coordinate transformation $e_i'=R_{iq}e_q$ for each factor of $e_i'$. This means you probably won't ever have two distinct transformations (as in your formula $T_{ij}'=A_{ip}B_{jq}T_{pq}$) since you always want to perform the same transformation to your basis in both factors of the tensor product $e_i\otimes e_j$.
