If we got the curve $\vec{\gamma}:[0,1]\rightarrow\mathbb{R^3}$
$$\vec{\gamma}(t) = \left(\! \begin{array}{c} t \\ t^2+1 \\ t \end{array} \!\right) $$
And the vector field $\vec{v}:\mathbb{R^3}\rightarrow\mathbb{R^3}$
$$\vec{v}(x,y,z) = \left(\! \begin{array}{c} f'(x) \\ g'(y) \\ h'(z) \end{array} \!\right) $$
Where $f,g,h:\mathbb{R}\rightarrow\mathbb{R}$ are continuous and differentiable.
What would $\vec{v}(\vec{\gamma}(t))$ become?
I'd say $$\vec{v}(t) = \left(\! \begin{array}{c} f'(t) \\ g'(t^2+1) \\ h'(t) \end{array} \!\right) $$ The composition gives you a vector depending only on the $t$ variabile.