It might be helpful to start with non relativistic hydrodynamics and ask for the change of the momentum of a fluid in time. Let $\rho$ be the mass density of an incompressible fluid. If $\rho v$ is the momentum per volume, then we can write:
$$\frac{\partial(\rho v)}{\partial t}=\rho \frac{\partial v}{\partial t}+\frac{\partial \rho}{\partial t} v.$$
The first term on the right hand side can be processed by the Euler equation:
$$\frac{\partial v^i}{\partial t}=-v^k \frac{\partial v^i}{\partial x^k}-\frac{1}{\rho}\frac{\partial p}{\partial x^i}$$
while $p$ is the pressure. The second term in the first equation can be replaced by the continuity equation:
$$\frac{\partial \rho}{\partial t}=-\frac{\partial (\rho v^k)}{\partial x^k}.$$
Then, after substitution of both and a few algebraic manipulations, we get:
$$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial p}{\partial x^i}-\rho\, v^k\frac{\partial v^i}{\partial x^k}-v^i\frac{\partial(\rho v^k)}{\partial x^k}.$$
Now, after applying the product rule on the right hand side we obtain the quadratic term of the velocity:
$$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial p}{\partial x^i}- \frac{\partial }{\partial x^k}(\rho\,v^iv^k).$$
The first term on the right hand side can be expressed with help of the Kronecker delta, i.e.:
$$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial }{\partial x^k}(\delta^{ik}p+\rho\,v^iv^k).$$
By applying the notation
$$T^{ik}:=\delta^{ik} p +\rho\,v^i v^k$$
we obtain the representation:
$$\frac{\partial(\rho v^i)}{\partial t}=-\frac{\partial }{\partial x^k}T^{ik}.$$
This result is easily generalized to the relativistic case.
