We know that the Moreau decomposition is normally wrote as,
$$x = \text{prox}_{h}(x) + \text{prox}_{h^*}(x)$$
But I can't really derive the more general form from the original one.
$$x = \text{prox}_{t,h}(x) + t\text{prox}_{\frac{1}{t},h^*}(x/t)$$
We already know that $(th)^*(x) = th^*(x/t)$, where $h^*(x)$ means the convex conjugate of $h(x)$
Simply apply the basic decomposition to the scaled function $th$ to get $$ \begin{aligned} x = \text{prox}_{th}(x) + \text{prox}_{(th)^*}(x) = \text{prox}_{th}(x) + \text{prox}_{t\tilde{h^*}}(x) = \text{prox}_{th}(x) + t\text{prox}_{\frac{1}{t}h^*}(x/t), \end{aligned} $$ where $\tilde{h^*}: y \mapsto h^*(y/t)$, so that $$ \begin{aligned} \text{prox}_{t\tilde{h^*}}(x) &:= \underset{y}{\text{argmin }}\frac{1}{2}\|y-x\|_2^2 + th^*(y/t) \\ &= \underset{y}{\text{argmin }}\frac{1}{2}\|y/t-x/t\|_2^2 + (1/t)h^*(y/t) \;\text{ ( dividing through by }t^2)\\ &=t \times \underset{z}{\text{argmin }}\frac{1}{2}\|z-x/t\|_2^2 + (1/t)h^*(z)\;\text{ ( using the change of variable }y = tz)\\ &=: t\text{prox}_{\frac{1}{t}h^*}(x/t). \end{aligned} $$