The angle between two vectors is always less than 180 degrees.
If we are talking about angles greater than 180 degrees we need some other vocabulary. We need an orientation. Which vector is "on the right."
One approach:
Construct a matrix with the two vectors you are comparing in the first two columns, and the rest of the columns as principal component vectors.
The sign of the determinant will tell you the orientation of the two vectors.
Update
In 3-D
Suppose your vectors are $(x_1,x_2,x_3), (y_1,y_2,y_3)$
Then the the sign of $\det \begin{bmatrix} x_1&y_1&0\\x_2&y_2&0\\x_3&y_3&1\end{bmatrix}$ will give you the orientation between your two vectors from the point of view of the positive z axis.
If you wanted to extend the dimension.
$\det \begin{bmatrix} x_1&y_1&0&0\\x_2&y_2&0&0\\x_3&y_3&1&0\\x_4&y_4&0&1\end{bmatrix}$
It's not apparent what angle is being discussed (after all, the twisted cubic lives in $\Bbb P^3$), or what is intended by "consider these angles represented as homogeneous coordinates," for example). Projective geometry has no invariant notion of angle (projective transformations change angles between curves, just as they change lengths${}^*$), so I'd suggest jettisoning that line of thinking as likely to mislead your intuition anyway.
I'd say that the description
...less $3$ for a $1$D projectivity on the parameterization $\theta$...
from the text is imprecise. What's going on here is the following:
(1) By way of analogy, recall that we can parameterize any (say, connected, piecewise $C^1$) curve $\gamma$ in Euclidean space by arc length---because these parameterizations respect the Euclidean structure (i.e., the notion of length) induced on the curve, they have special status among all parameterizations of $\gamma$. On the other hand, arc length parameterizations are not unique: If $\gamma(s)$ is one such parameterization, so is $\gamma(s + A)$ for any constant $A$. On the other hand (if $\gamma$ is oriented) all (oriented) parameterizations arise this way, so our parameterization has exactly one degree of freedom.
There is an analogous description for curves $\gamma$ in projective geometry. This time, projective space induces a projective structure on the curve itself, and while $\gamma$ has many parameterizations, only some of them respect this structure. In this case, if $\gamma(\theta)$ is one such parameterization, the others are precisely the parameterizations $$\theta \mapsto \gamma\left(\frac{A \theta + B}{C \theta + D}\right),$$ where $A D - B C \neq 0$---we call these projective parameterizations or projective changes of variable (or often outside the context of projective geometry, linear fractional transformations). But $(A, B, C, D)$ and $(\lambda A, \lambda B, \lambda C, \lambda D)$ define the same parameterizations, so projective reparameterizations only entail $3$ degrees of freedom.
We can of course make this explicit for our twisted cubic: If our original parameterization is $${\bf X}(\theta) = [X^i(\theta)] = [a_{i1} + a_{i2} \theta + a_{i3} \theta^2 + a_{i4} \theta^3] ,$$ then applying a projective transformation gives the new parameterization
\begin{align}
{\bf X}\left(\frac{A \theta + B}{C \theta + D}\right)
= &\left[X^i\left(\frac{A \theta + B}{C \theta + D}\right)\right] \\
= &\left[a_{i1} + a_{i2} \left(\frac{A \theta + B}{C \theta + D}\right) + a_{i3} \left(\frac{A \theta + B}{C \theta + D}\right)^2 + a_{i4} \left(\frac{A \theta + B}{C \theta + D}\right)^3\right] \\
= &\left[a_{i1} (C \theta + D)^3 + a_{i2} (A \theta + B) (C \theta + D)^2\right. \\
&\qquad \left. + a_{i3} (A \theta + B)^2 (C \theta + D) + a_{i4} (A \theta + B)^3\right]
\end{align}
We can collect like terms in $\theta$ to produce a new parameterization of the same twisted cubic of the form
$${\bf X} = {\bf B} [\theta^i]$$
for some matrix $\bf B$ whose entries are linear in the entries $a_{ij}$.
Since $3$ of our $15$ degrees of freedom only change the parameterization and not the choice of twisted cubic, we expect that it takes $15 - 3 = 12$ real parameters to specify a twisted cubic. Any point specifies $2$ real parameters, so generically we expect that it takes $\frac{12}{2} = 6$ points to specify a twisted cubic. (You might already be familiar with the fact that it takes only $5$ points to specify a conic in $\Bbb P^2$.)
(2) Here, "eliminating" $\theta$ just means eliminating the parameter $\theta$ in a system of equations. Analogously, with some algebra we can describe an affine line in $\Bbb R^2$ given by a parameterization $t \mapsto (v t + x, w t + y)$ with an equation of the form $a x + b y + c = 0$, which in particular does not involve $t$. Notice that do not actually need to solve for the parameter explicitly to carry out the relevant counting.
Remark It's natural to ask why the linear fractional transformations are exactly the reparameterizations that respect the projective structure. We can, for example, view any projective parameterization of the twisted cubic as an embedding $\theta : \Bbb P \to \Bbb P^3$. In particular, if we apply a projective transformation $T$ to $\Bbb P^3$ that fixes a particular cubic, then $\theta^{-1} \circ T \circ \theta : \Bbb P \to \Bbb P$ is a projective transformation of $\Bbb P$, but we know what these transformations are: They are given in homogeneous coordinates exactly by the action of $PGL(2, \Bbb R)$, which in any affine chart on $\Bbb P$ acts by linear fractional transformations.
${}^*$ In fact there is a projectively invariant notion of arc length, but it is rather different in flavor from the familiar Euclidean one---which is emphatically not preserved by projective transformations. If you're interested in computational aspects of projective invariants like this one, I highly recommend Clelland's From Frenet to Cartan: The Method of Moving Frames, though you should have some elementary differential geometry under your belt (curves and surfaces in Euclidean space, vector fields, $1$-forms, exterior derivative) before taking it up.
Best Answer
To define a line in $\mathbb{R}^n$, we need two points. This gives $2n$ degrees of freedom.
But for one line each point can be any point in the line. This substracts one degree of freedom for each point, so we have $2n-2$ degrees of freedom.
Another way: in $\mathbb{R}^n$ the line direction has as many degrees of freedom as (half of) the unit sphere $S^n$, i.e. $n-1$ because it is an hyper-surface.
Then the set of lines that have a given direction is in bijection with an hyperplane orthogonal to that direction. This adds $n-1$ degrees of freedom, so total is $2n-2$.