I am following these notes.
While introducing projective transformations (first sections of 2.3), the author seems to suggest that $T: V \to W$ need be invertible for $\tau: P(V) \to P(W)$ to be "well-defined". I interpret this to mean $\tau$ should take all representatives of say $x \in P(V)$ to the same point in the image of $\tau$.
However, it seems like I can construct a simple $T$ that induces a $\tau$ that is not surjective over $P(W)$ but is nonetheless well-defined. Consider $T: V \to W$ where $\dim V = 2$ and $\dim W = 1$. $V$ and $W$ generated by $\{ v_1, v_2 \}$ and $\{w_1\}$ respectively. If $\tau([v_1]) = [w_1]$ and $\tau([v_2]) = 0$, this seems to me to be a "projective transformation" (a well-defined map between projective points) where $T$ is not invertible.
Is there an implicit understanding that a projective transformation is another word for a bijective map in these notes? If so, why is that not more clear?
So-called homogeneous coordinates of "projective points" (in most definitions that I've seen) need to have at least one coordinate be nonzero. See the discussion of "representative points" on page 5 of those notes: the notation [0] doesn't even make sense.
If you have an invertible $T$, you know that $T(0) = 0$, so you can restrict $T$ to the nonzero elements of $V$ to get a new function $T'$, and (because $T$ is invertible) the image of $T'$ will be the nonzero elements of $W$. And because $T$ takes lines through the origin to lines through the origin, $T'$ takes punctured-lines-through-the-origin to punctured-lines-through-the-origin, rather than sending some of them to the $0$-dimensional subspace consisting of just the origin.
To put all this in the terms fo the notes you're reading:
For Hitchin, a projective point is a 1-d subspace of your vector space. If you look at real 2-space, and the function $T(x, y) = (0, y)$, then you'll see that the $x$-axis (a typical 1-d subspace!) is sent to the set $\{(0, 0)\}$ by this transformation, and that's a 0-d subspace. So this particular $T$ cannot be a projective xform. And in general, if you have a linear transformation that sends some 1-D subspace to a 0-D one, then (a) T is not a bijection, (b) T is not invertible, and (c) T sends some projective point to the "zero" subspace. (Indeed, these three conditions are equivalent.)