How to "reach" on various Tensors on Physics starting in the second tensor form? Well, I think that it will be a silly question but I have a strong doubt in how to "see" Tensors on Physics and "see beyond" the canonical formulation to modelate with tensors. I know the definitions, physical motivations etc...But since we have a canonical form of second rank tensors:
\begin{equation}
T_{ij} = \frac{\partial x^i}{\partial y^l}\frac{\partial x^j}{\partial y^k}T_{lk}
\end{equation}
How can I reach the inertia tensor, for example? Or the electromagnetic tensor? Metric tensor? They're all second rank tensors!
Just to show: 
Inertia Tensor
\begin{equation}
I_{ij} = \int \rho[r^2\delta_{ij}-r_{i}r_{j}]dV
\end{equation}
Eletromagnetic Tensor
\begin{equation}
F_{ij} = \frac{\partial A_i}{\partial x^j}-\frac{\partial A_j}{\partial x^i}
\end{equation}
Obs: Can I equal like:
\begin{equation}
T_{ij} = \frac{\partial x^i}{\partial y^l}\frac{\partial x^j}{\partial y^k}T_{lk} = I_{ij} = \int \rho[r^2\delta_{ij}-r_{i}r_{j}]dV
\end{equation}
Thank you all.
 A: I am not sure what is the question exactly. Probably there is a misunderstanding in the concepts.
A second rank tensor (i will call it a tensor simply) is a map $\mathfrak{T} :T\mathcal{M}\times T\mathcal{M}\rightarrow\mathbb{R}$ s.t. takes two vectors of a manifold and associates to the a single number with the condition of linearity; i.e. $\mathfrak{T}(a\mathbb{v}+b\mathbb{w},u) = a\mathfrak{T}(\mathbb{v},u)+b\mathfrak{T}(\mathbb{w},u)$. From  this and with the transformation law of a vector you can arrive to the fact that if $\mathfrak{T}$ is a tensor then, its components change in a determined way when we change the coordinates, as you pointed:
$$\mathfrak{T}_{ij} = {\Lambda_i}^k{\Lambda_j}^l\mathfrak{T}_{kl}\quad \mbox{with}\quad{\Lambda_a}^b=\frac{\partial y^b}{\partial x^a}$$
In this sense, this serves as a check for tensors: if a tensor is it truly, then should satisfy that expression. 
However you are starting from the oposite side: you state that the electromagnetic tensor is a tensor, then the formula can be used to calculate the components in other coordinates. 
Lets restate the question: Is this expression $F_{ij}=\frac{\partial A_j}{\partial x^i}-\frac{\partial A_i}{\partial x^j}$ a tensor, knowing that $A_i$ (in abstract index notation) is a vector? 
Let's preform a change of coordinates. Because $A_i$ is a vector, then it should transform as $A_i = \frac{\partial y^m}{\partial x^i}A'_m$. Then:
$$
F_{ij}=\frac{\partial A_j}{\partial x^i}-\frac{\partial A_i}{\partial x^j}=\frac{\partial}{\partial x^i}\left(\frac{\partial y^m}{\partial x^j}A'_m\right)-\frac{\partial}{\partial x^j}\left(\frac{\partial y^n}{\partial x^i}A'_n\right) \\
=\frac{\partial^2y^m}{\partial x^i\partial x^j}A'_m+\frac{\partial y^m}{\partial x^j}\frac{\partial A'_m}{\partial x^i}-\left(\frac{\partial^2y^n}{\partial x^j\partial x^i}A'_n+\frac{\partial y^n}{\partial x^i}\frac{\partial A'_n}{\partial x^j}\right)\\ =
\frac{\partial^2y^m}{\partial x^i\partial x^j}A'_m+\frac{\partial y^m}{\partial x^j}\frac{\partial y^n}{\partial x^i}\frac{\partial A'_m}{\partial y^n}-\left(\frac{\partial^2y^n}{\partial x^j\partial x^i}A'_n+\frac{\partial y^n}{\partial x^i}\frac{\partial y^m}{\partial x^j}\frac{\partial A'_n}{\partial y^m}\right)
$$ 
Where the chain rule has been used. Because $\frac{\partial}{\partial x^i\partial x^j} = \frac{\partial}{\partial x^j\partial x^i}$ and the indices $m,n$ are dummy, the first and third terms cancell each other. Rearranging the terms:
$$
F_{ij} = \frac{\partial y^m}{\partial x^j}\frac{\partial y^n}{\partial x^i}\frac{\partial A'_m}{\partial  y^n}-\frac{\partial y^n}{\partial x^i}\frac{\partial y^m}{\partial x^j}\frac{\partial A'_n}{\partial  y^m}=
\frac{\partial y^m}{\partial x^j}\frac{\partial y^n}{\partial x^i}\left(\frac{\partial A'_m}{\partial  y^n}-\frac{\partial A'_n}{\partial  y^m} \right)=\frac{\partial y^m}{\partial x^j}\frac{\partial y^n}{\partial x^i}F'_{mn}
$$
So indeed $F$ is a tensor. Recall that we used this to check that it is a tensor. The formula just describeas the transformation law for every (2-rank covariant) tensor. The formula does not relates to any specific tensor, or better: it relates to every tensor. Therfor, you can allways write the last equation you wrote, you are just changing the name if the tensor $I$ for $T$
