(Expanding on a comment by John Ma.)
The family is uniformly bounded: $|f_n|\le M_0$. It is also equicontinuous, thanks to $|f'_n|\le M_1$:
$$
|f_n(x)-f_n(y)|\le M_1|x-y|
$$
Therefore, there is a subsequence $\{f_{n_k}\}$ converging to a continuous function $f$.
Next step, consider $\{f_{n_k}'\}$ which is also a uniformly bounded equicontinuous family. Extract a further subsequence $f'_{n_{k_l}}$ which converges uniformly. Since $f_{n_{k_l}}\to f$ and $f'_{n_{k_l}}$ converge uniformly, it follows that $f$ is differentiable and $f'_{n_{k_l}}\to f'$.
This process of choosing subsequences continues indefinitely, demonstrating that $f$ is infinitely differentiable. It remains to apply the diagonal argument: let $F_n$ be the $n$th term of the $n$th subsequence. Then $F_n$ converge, with all its derivatives, to the corresponding derivatives of $f$.
If $v$ is zero at some point, then this means that the 'velocity' of the curve is $=0$ In such a point the image of the parametrization may have all kinds of singularities, which -- in a geometric context -- is not desired.
As a simple example you may consider the curve
$$c(t) = (t^5, |t|^{5/2})$$
which is (quite obviously) $C^1$, but which is a parametrization of the graph of $y= \sqrt{|x|}$, which has a cusp at $t=0$. Of course this kind of example can be given in any dimension.
With the help functions like $\exp(-\frac{1}{|x|^2})$ it's possible to create parametriziations of such curves which are actually smooth.
The requirement $v\neq 0$ prevents the curve from having geometric singularities, for this reason one calls a curve with this property regular.
Edit (in response to a question in a comment): if you want to define line integrals then it is not really necessary to assume that the curves are regular. With this assumption the definition is, however, much easier and straightforward. Just assuming that there is a smooth parametrization, is, for example, not sufficient.
The example I gave is still a 'nice' curve and would allow do define a line integral along the curve without much of a problem. One can write down more nasty ones so that the parametrization is still smooth, but the curve no longer has finite length or has other nasty features like fractal behavior.
This does not automatically mean that the definitions of length and line integral don't make any sense anymore, it may just be more difficult to give a consistent defintion.
The case of corners is often the first generalization which is introduced as admissible, since that class of curves is often more convenient in applications.
Best Answer
First fact: every rational function is differentiable and its derivative is a rational function.
This is clear since
$$\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x) g(x) - f(x) g'(x)}{g^2(x)}$$
Iterating the argument, you can show that every rational functions is infinitely differentiable (formally you have to use induction).