Let $S$ and $T$ be spheres with radius $R_S$ and $R_T$ respectively. Define the diffeomorphism $\Phi:S \to T$ by $\Phi(s) = \frac{R_T}{R_S}s$. Given a function $u:T \to \mathbb{R}$, we can define $\tilde u:S \to \mathbb{R}$ by $$\tilde u(s) = u(\Phi(s)) = u\left(\frac{R_T}{R_S}s\right).$$
I am trying to write the Laplace-Beltrami operator $\Delta_T$ of $u$ in terms of $\Delta_S$ and $\tilde u$. I start with the formula
$$\Delta_{S(R)} f(x) = R^2\Delta f\left(R\frac{x}{|x|}\right)\tag{1}$$ which holds for any sphere $S(R)$ of radius $R$.
Using this I see that $$\Delta_S \tilde u(s) = R_S^2\Delta \tilde u\left(R_S\frac{s}{|s|}\right) = R_S^2\Delta u\left(R_T\frac{s}{|s|}\right)$$ with the last equality by definition of $\tilde u$. I want to say this is equal to $$\frac{R_S^2}{R_T^2}\Delta_T u(s)$$ by the formula (1). But then since $s \in S$ it doesn't make sense to write the above expression since we are taking the Laplace-Beltrami over $T$. And I'm not sure that I can use (1) because in (1) $x \in S(R)$, and here $s \in S$ and not $s \in T$.
How do I do this calculation correctly??
Indeed the expression $\Delta_T\, u(s)$ doesn't make sense when $s \in S$.
To understand what to replace it with, notice the definition of the Laplace-Beltrami operator on spheres, which you quoted in formula (1).
The function $f$ on the left-hand side can be thought of as a function that's only defined on the sphere of radius $R$.
On the right side, we take the regular (Euclidean) Laplacian. This is NOT the Laplacian of $f$. It is the Laplacian of the spherical-extension of $f$, which is the function $$x \mapsto f\left(R \frac x {\left|x\right|}\right)$$
This can be a bit confusing. When you write $\Delta f\left(R \frac x {\left|x\right|}\right)$, it might appear as if you're taking the regular Laplacian of $f$ and plugging in $R \frac x {\left|x\right|}$ (which is absurd, as $f$ is not even defined in any open set of the Euclidean plane, so it has no Euclidean Laplacian!). It might be more instructive (but more cumbersome) to write instead $\Delta\left(x \mapsto f\left(R \frac x {\left|x\right|}\right)\right)$, or to define an auxiliary function (the extension) of which the Laplacian will be taken.
Now, what properties does this extension of $f$ have?
Back to your problem: To simplify the calculations I'll introduce new notations to the extensions of $u$ and $\tilde u$ to the entire plane (minus the origin), and I'll call them $U$ and $\tilde U$:
$$U(x) = u\left(R_T \frac x {|x|}\right)$$ $$\tilde U(x) = \tilde u\left(R_S \frac x {|x|}\right)$$
But notice that $U$ and $\tilde U$ are actually exactly the same function! Follow their definitions (and the "equality" of $u$ and $\tilde u$) to see why this is true.
By definition of the Laplace-Beltrami operator, for any $s \in S$ and $t \in T$:
$$\Delta_T\, u(t) = R_T^2\, \Delta U(t)$$ $$\Delta_S\, \tilde u(s) = R_S^2\, \Delta U(s)$$
To proceed, we need to study the function $\Delta U$ and see how it behaves with respect to "stretching" the parameter. That is, given $\Delta U(x)$, what is $\Delta U(a x)$ for $a > 0$?
Notice that, by definition, $U(x) = U(ax)$ for $a > 0$. By differentiating both sides of this equation twice and using the chain rule, we get:
$$\frac {\partial^2} {\partial x_i^2} U(x) = a^2 \frac {\partial^2} {\partial x_i^2} U(ax)$$
Adding up all the partial second derivatives we get:
$$\Delta U(x) = a^2 \Delta U(ax)$$
for any $a > 0$.
Finally, using all of the above, we can make the calculation you wanted. Let $t \in T$ be some value and let $s \in S$ be defined by $s = \frac {R_S} {R_T} t$. Then:
$$\Delta_T\, u(t) = R_T^2\, \Delta U(t) = R_T^2\, \Delta U\left(\frac {R_T}{R_S} s\right) = R_S^2\, \Delta U(s) = \Delta_S\, \tilde u(s)$$
The conclusion: Thanks to the normalization in the definition you used of the Laplace-Beltrami operator (the multiplication by the radius squared), the value of the Laplacian does not depend on the radius of the sphere on which any given spherical function is defined.