I'm having some trouble on understanding how to calculate the derivative of $g(t)$ with regards to $t$, where $g(t) := f(X(t), Y(t))$ and $X(t)$ and $Y(t)$ are $2d$ vectors. That is $X,Y: R \to R^2$. What would the chain rule say in this case?
It's very straightforward if $X,Y: R \to R$ and f is a function of two variables. Then we just have $\frac{d g}{ d t} = \frac{\partial f}{\partial X}\frac{\partial X}{\partial t} + \frac{\partial f}{\partial Y}\frac{\partial Y}{\partial t}$. But now, $f$ is a function of two $2d$ vectors (would that be $f: R^4 \to R$ ??) and $X,Y: R \to R^2$.
If it helps, you may write : $\forall t \in \mathbb{R}, \, X(t) = \begin{bmatrix} x_{1}(t) & x_{2}(t) \end{bmatrix}^{\top} \quad \text{and} \quad Y(t) = \begin{bmatrix} y_{1}(t) & y_{2}(t) \end{bmatrix}$. So that :
$$ \forall t \in \mathbb{R}, \, g(t) = f\big( x_{1}(t),x_{2}(t),y_{1}(t),y_{2}(t) \big). $$
The chain rule writes :
$$ \small{ g'(t) = x_{1}'(t) \frac{\partial f}{\partial z_{1}}\big( X(t),Y(t) \big) + x_{2}'(t) \frac{\partial f}{\partial z_{2}}\big( X(t),Y(t) \big) + y_{1}'(t) \frac{\partial f}{\partial z_{3}}\big( X(t),Y(t) \big) + y_{2}'(t) \frac{\partial f}{\partial z_{4}}\big( X(t),Y(t) \big). }$$