Let $V =(v_{1} \wedge \ldots \wedge v_{k})$ be a smooth curve in the $k$-th exterior power $\Lambda^{k}(\mathbb{R}^{n})$ of $\mathbb{R}^{n}$.
In the case $n=3$ and $k=2$, I believe the following holds: $$\frac{dV}{dt} = \frac{dv_{1}}{dt} \wedge v_{2} + v_{1} \wedge \frac{dv_{2}}{dt}.$$
I would like to know how the above relation generalizes to the case $k=n-1$. For example, when $n=4$ and $k=3$, is that true that $$\frac{dV}{dt} = \frac{dv_{1}}{dt} \wedge v_{2} \wedge v_{3}\ + v_{1} \wedge \frac{dv_{2}}{dt} \wedge v_{3} + v_{1} \wedge v_{2} \wedge \frac{dv_{3}}{dt}?$$
The generalization you expect is correct and it works for any $k$ and $n$. What you need in order to prove this is just the fact how to differentiate multilinear maps between vector spaces, and apply this to the map $(\mathbb R^n)^k\to\bigwedge^k(\mathbb R^n)$ defined by $(v_1,\dots,v_k)\mapsto v_1\wedge\dots\wedge v_k$.
Edit: If you have a multilinear map $\Phi:V_1\times\dots\times V_k\to V$, then the derivative $D\Phi(v_1,\dots,v_k)(w_1,\dots,w_k)$ (with $v_i,w_i\in V_i$) can be computed as a directional derivative. So you consider the curve $\Phi(v_1+tw_1,\dots,v_k+tw_k)$, differentiate with respect to $t$ and evaluate at $t=0$. Using linearity in each component, you can expand $\Phi(v_1+tw_1,\dots,v_k+tw_k)$ in powers of $t$, and you want to find the coefficient of $t$. This is just the sum of all terms in which one $v$ is replaced by the corresponding $w$, i.e. $\Phi(w_1,v_2,\dots,v_k)+\dots+\Phi(v_1,\dots,v_{k-1},w_k)$.