What does it mean to say "tensors transform sensibly"? I have been reading about tensors and they are described as "objects which transform in a physically meaningful/sensible manner" and obey the equation (for rank 2, but generalises) $T'_{ij}=R_{ip}R_{jq}T_{pq}$. I am struggling to understand why this equation makes $T_{ij}$ that it is physically meaningful. I think part of my misunderstanding comes from not getting what $T_{ij}'$ is. If I change basis then the tensor $T_{ij}$ becomes $T_{ij}'$. Well what if the change of basis isn't an orthogonal matrix? Then for any tensor $T_{ij}'\neq R_{ip}R_{jq}T_{pq}$ But if the change of basis matrix is orthogonal, $M_ij'=R_{ip}R_{jq}M_{pq}$ for any matrix! May I ask that an answer includes a concrete example. Thanks
I tried reading wikipedia and some other answers here, but they often use other definitions not covered in a physics course.
 A: Although it is difficult, I will try to avoid mathematical definitions here. A tensor $T$ is, first and foremost, a geometric object. It lives in a space on its own and bears no reference to anything else.
When we choose basis vectors, the tensor $T$ can be expressed as a unique linear combination of the basis. In your example, it is
$$T=T_{kl} \mathbf{e}^k \otimes\mathbf{e}^l$$
The coefficients $T_{kl}$ are known as the components of $T$ in this particular basis. Now, let's choose another basis $\bar{\mathbf{e}}^i = M^i_j \mathbf{e}^j$ where $M$ is any invertible matrix. It must be true that we can also express $T$ as a linear combination of the new basis
$$T=\bar{T}_{ij} \bar{\mathbf{e}}^i \otimes \bar{\mathbf{e}}^j$$
where $\bar{T}$ are the new components. Substituting it into the above equation, we get
$$T=\bar{T}_{ij} \bar{\mathbf{e}}^i \otimes \bar{\mathbf{e}}^j = \bar{T}_{ij} M^i_k \mathbf{e}^k \otimes M^j_l \mathbf{e}^l = \bar{T}_{ij} M^i_k M^j_l \mathbf{e}^k \otimes \mathbf{e}^l$$
from which we see that $T_{kl} = \bar{T}_{ij} M^i_k M^j_l$ by comparing with our first equation. We have derived the tensor transformation law by requiring nothing other than the fact that $T$ itself remains unchanged. The only other condition needed is that the transformation matrix is invertible. You can also check that inverting the transformation gives
$$\bar{T}_{ij} = T_{kl}\left(M^{-1}\right)^k_i \left(M^{-1}\right)^l_j$$
This is very significant in physics because the laws of physics do not depend on the choice of coordinate system.
A: 
May I ask that an answer includes a concrete example.

In physics, we actually consider tensor fields like the electromagnetic field tensor. To understand tensor fields, you first need some knowledge of multilinear algebra, manifolds and tangent spaces.
However, it is not difficult to see where your transformation rule originates: Given some finite-dimensional vector space, each basis defines an isomorphism between tensors and matrices and your formula is the transformstion rule for the matrices associated to the same tensor through different bases.
Derivation of the transformation rule
Let $V$ be an $n$-dimensional vector space over a field $F$ and $I:=\{1,\ldots,n\}$. Then the dual space
$$V^*:=L(V,F)$$
is the set of all linear functions from $V$ to $F$. Let $v_1,\ldots,v_n$ be a basis of $V$, then $$v^1,\ldots,v^n\in V^*$$ is a basis of $\displaystyle{V^*}$ - the dual basis. It follows from the universal property of the tensor product that
$$\{v^k\otimes v^l:k,l\in I\}\subset V^*\otimes V^*$$
is a basis of $\displaystyle{V^*\otimes V^*}$, the vector space of rank $2$ tensors mentioned in your question. That is, the function
\begin{align}
F^{n\times n}&\to V^*\otimes V^*\\
A&\mapsto A_{kl}\cdot v^k\otimes v^l
\end{align}
is a bijection, where $\displaystyle{F^{n\times n}}$ denotes the set of all $n\times n$-matrices with entries in $F$.
Now let $w_1,\ldots,w_n\in V$ be another basis and consider the matrix $R\in\displaystyle{F^{n\times n}}$ defined by
$$R^k{}_l:=v^k(w_l)$$
for all $k,l\in I$. Suppose $t\in\displaystyle{V^*\otimes V^*}$ and
$$t=T_{kl}\cdot v^k\otimes v^l={T'}_{kl}\cdot w^k\otimes w^l,$$
then it can be shown that
$${T'}_{ij}=R^k{}_iR^l{}_jT_{kl},$$
for all $i,j\in I$, which is your transformation rule.
