The concrete problem is this
Let $$u(t,x)=v(\frac{x^2}t)$$ on $\mathbb R^+\times\mathbb R$.
Show that
$$u_t=u_{xx} \Leftrightarrow 4zv''(z)+(z+2)v'(z)=0$$
Now, while I would like to know the solution to this problem, I'm more interested in the general approach. The way I interpret this substitution is that it says the function $u$ is constant along any parabole $t=cx^2$. But how do I turn this knowledge into something I can work with? What does $v'$ mean in that context? If I do this substitution, does it then make sense to talk about $u_z$? What does THAT mean?
Set $z=\frac{x^{2}}{t}$ Then, by the ordinary chain rule
$u_{t}(x,t)=-\frac{x^{2}}{t^{2}}v'(z)$. Recalling what $z$ is, we can write this as
$\bf u_{t}(t,x)=-\frac{z}{t}v'(z)$
On other hand, by the chain and product rules
$u_{x}(t,x)=2\frac{x}{t}v'(z)\\u_{xx}(t,x)=\frac{2}{t}v'(z)+4\frac{x^{2}}{t^{2}}v''(z)$.
Recalling what $z$ is, this last item becomes
$\bf u_{xx}(x,t)=\frac{2}{t}v'(z)+4\frac{z}{t}v''(z)$.
Now just inspect the highlighted items.
Comment: $u$ is a map: $R^{+}\times R\rightarrow R$ and is evidently equal to the composition $g:R^{+}\times R\rightarrow R$ (given by $g(t,x)=\frac{x^{2}}{t}$) and $v:R\rightarrow R$. That is, $u=v\circ g$.
The definition of partial derivative then allows us to fix one of the coordinates and differentiate with respect to the other in the "usual" way, so that , for example, if we want $u_{x}$ at the point $(a,b)\in R^{+}\times R$, we simply compute $(v\circ g)_{x}(a,b)$ which, using the ordinary chain rule (t is fixed remember), becomes $v'(g((a,b))g_{x}(a,b)=v'\left ( \frac{a^{2}}{b} \right )g_{x}(a,b)$ where we are allowed to use $v'$ because $v$ is a function on $R$.