Consider a parametric smooth surface: $$S:R^2 \to R^3, \quad S(u,v)=(X_S(u,v),Y_S(u,v),Z_S(u,v)), \quad for \ u,v \in R$$ a parametric smooth plane curve: $$q:R \to R^2, \quad q(t)=(X_q(t),Y_q(t)), \quad for \ t \in R$$ and the space curve: $$c:R \to R^3, \quad c(t) = (S\circ q)(t)$$ Now, I believe that the following hold: $$\frac{dq}{dt}\in R^2 \ , \ \frac{\partial{S}}{\partial{u}},\frac{\partial{S}}{\partial{v}} \in R^3, \quad \frac{dc}{dt} \in R^3 $$ $$\frac{dc}{dt} = \left. \frac{dS(u,v)}{dt}\right\vert_{(u,v)=q(t)}* \frac{dq(t)}{dt} $$
So obviously something is wrong. What kind of multiplication is $R^3*R^2$?
Is$\left. \frac{dS(u,v)}{dt}\right\vert_{(u,v)=q(t)} \notin R^3$?
You are multiplying Jacobian matrices, which are the higher dimensional analogs of the derivative for maps $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$. In your notation, $dq/dt$ is the $2\times 1$ Jacobian matrix of $q$:
$$\begin{bmatrix}\frac{dX_q(t)}{dt}\\ \frac{dY_q(t)}{dt}\end{bmatrix}$$
In your notation, concatenating $\partial S/\partial u$ and $\partial S/\partial v$ into a matrix we obtain the Jacobian of the map $S$:
$$\begin{bmatrix}\frac{\partial X_S(u,v)}{\partial u}&\frac{\partial X_S(u,v)}{\partial v}\\ \frac{\partial Y_S(u,v)}{\partial u}&\frac{\partial Y_S(u,v)}{\partial v}\\ \frac{\partial Z_S(u,v)}{\partial u}&\frac{\partial Z_S(u,v)}{\partial v}\\ \end{bmatrix}$$
You'll find that the product of these two matrices will give you a higher dimensional analog of the chain rule. You should get the corresponding $3\times 1$ Jacobian for the composed map.