Substantial / Total / Material Derivative Contradiction We want to find the material derivative in terms of the spatial representation (classic case).
$\phi$ : Temperature, entropy etc,
$x_{0}$ : Reference coordinate,
$x$ : Spatial / eulerian coordinate
Let $\phi=xt$ and $x= x_{0}(1+t)$ where $x_{0} = c $
$\dfrac{d\phi}{dt} = \left(\dfrac{\partial \phi}{\partial x}\right)_{t} \dfrac{dx}{dt}+ \left(\dfrac{\partial \phi}{\partial t}\right)_{x}$ 
Therefore, the total derivative in spatial and material coordinates is the following
$\dfrac{d\phi}{dt} = tx_{0} + x$
To obtain the pure spatial representation, you find the material coordinate in terms of the spatial coordinate.
$x_{0} = \dfrac{x}{1+t}$
Lastly, you perform the substitution into the total derivative yielding the classic result.
$\dfrac{d\phi}{dt} = t\dfrac{x}{1+t} + x$
Since $x_{0} = c$,  how can the above substitution be valid for any $x, t$ when it appears only valid for $x(t)= x_{0}(1+t)$? 

The material derivative is the temporal rate of change of the scalar $\phi$ following the path of a fluid particle.  In a Lagrangian framework, the location of a fluid particle at time $t$ initially with coordinate $x_0$ is specified by a smooth function $\xi: \mathbb{R}^2 \to \mathbb{R}$ where 
$$x = \xi(x_0,t) = x_0(1 + t).$$
Each fluid particle is associated with a different $x_0$ which you are mistakenly interpreting as a fixed constant.
The material derivative is
$$\frac{d}{dt} \phi(\xi(x_0,t),t) = D_1\phi(\xi(x_0,t),t) \frac{\partial \xi}{\partial t} + D_2\phi(\xi(x_0,t),t), $$
where $D_1$ and $D_2$ denote partial derivatives with respect to the first and second arguments.  (Part of you confusion may be due to your notation).
This reduces to 
$$\frac{d}{dt} \phi(\xi(x_0,t),t)  =tx_0 + \xi(x_0,t), $$
and written in terms of Eulerian coordinates
$$\frac{d \phi}{dt}   =t\frac{x}{1+t} + x. $$
