I currently understand that a homography matrix, which allows for a mapping between planes in 3-dimensions, is a $3\times3$ matrix of the following general form:
$$\begin{bmatrix} \vert & \vert & \vert \\ r_1 & r_2 & t \\ \vert & \vert & \vert \end{bmatrix}$$
In my class, we were introduced to the idea that a 2d homography has special cases including similarity transformations, affine transformations, and projective transformations.
Furthermore, the components of the $H$ matrix were broken up into four "quadrants".
$$H=\left[\begin{array}{c|c} \\[-1ex] \quad A\quad & t \\[-1ex]\\\hline V & v \end{array}\right]$$
Would someone be able to relate these transformations to the various elements in $H$?
For a similarity matrix, my understanding is that the image can only be rotated in the $xy$ plane, scaled, and translated. Thus a similarity $H$ matrix would have zeros for elements $h_{31}$ and $h_{32}$ and $1$ for $h_{33}$.
I am unclear how to interpret affine and projective transformations in this way, especially because their definitions are not as clear to me.
Any matrix, usually excluding singular matrices (determinant zero). That's the most general case.
Preserve points at infinity, i.e. result has $z=0$ if and only if input has $z=0$. That is what you describe with $h_{31}=h_{32}=0$ and $h_{33}=1$, except that last coordinate $h_{33}$ can in fact be any non-zero value since a multiple of that matrix describes the same transformation. So you'd have $V=0$ and $v\neq0$.
It makes sense to view an affine transformation algebraically, on inhomogeneous coordinates:
$$x'=(h_{11}x + h_{12}y + h_{13})/h_{33}\\ y'=(h_{21}x + h_{22}y + h_{23})/h_{33}$$
Here you can see that $v$ serves as a global denominator, $t$ describes a translation and $A$ a linear 2d transformation.
These are affine transformations which also preserve angles. For a rotation, the upper left $2\times2$ block $A$ would be a rotation matrix like $\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$. But for a similarity these may be scaled as well, so you'd get something of the form $A=\begin{bmatrix}a&-b\\b&a\end{bmatrix}$, together with $V=0$ and $v\neq 0$.