Original PDE;
$$ \frac{\partial T}{\partial r} = \kappa \frac{1}{r^2} \frac{\partial}{\partial r} \big( r^2 \frac{\partial T}{\partial r} \big), \quad 0 \leq r \leq R, \quad t \geq 0 $$ IC: $ \quad T(r,0)=T_0, \quad $ BC: $\quad \frac{\partial T}{\partial r}(0,t)=0, \quad k\frac{\partial T}{\partial r}(R,t)=-\beta \big( T(R,t)-T_1 \big)$
When scaling to dimensionless form, I use these variables;
$$ x = \frac{r}{R}, \quad u=\frac{T(r,t)-T_1}{T_0-T_1}, \quad \tau=\frac{t}{\alpha} $$
When I put in these three variables into the original PDE, I find that $\alpha = \frac{\kappa}{R^2}$ if I want the PDE to become dimensionless. It now takes the form $\frac{\partial u}{\partial \tau}=\frac{1}{x^2} \frac{\partial}{\partial x} \big( x^2 \frac{\partial u}{\partial x} \big)$.
My question is regarding the boundary/initial conditions. When I derive the new dimensionless PDE, every instance of $T$ is replaced with $T_1 + (T_0 - T_1)u$, every instance of $t$ is replaced with $\tau \alpha$, and every instance of $r$ is replaced with $Rx$.
However, when I want to derive the new initial condition, I write $T(r,0)=T_0 \implies T_1 + (T_0 - T_1)u(x,0)=T_0 \implies u(x,0)=1$. My question is, why should I write $u(x,0)$ and not $u(Rx,0)$, since $r=Rx$?
You are making the change of variables $$u(x,\tau) := \frac{T(R x ,\alpha \tau) - T_1}{T_0 - T_1}$$ This means that $$u(x,0) = \frac{T(R x, 0) - T_1}{T_0 - T_1} = 1$$ since $T(r,0) = T_0$ for any $r$ in particular $r = Rx$. And $$\partial_r = R\partial_x, \quad \partial_t =\alpha \partial_\tau$$ so for example $\partial_t T(r,t) = \frac{1}{\alpha}\partial_\tau T(Rx,\alpha \tau) = \frac{1}{\alpha}\partial_\tau u$. All together we have $$\frac{1}{\alpha} \partial_\tau u = \frac{\kappa R^2}{x^2}\frac{1}{R}\partial_x ((\frac{x}{R})^2 \frac{1}{R}\partial_x u)$$ $$\partial_\tau u = \frac{\alpha \kappa}{ R^2}\frac{1}{x^2}\partial_x (x^2 \partial_x u)$$ so $\alpha = R^2/\kappa $. This makes sense as $\kappa$ should have units of distance squared over time.