From what I've learnt, since I'm also a student, is that you want to work with $4 \times 4$ matrices in order to treat rotations, scaling and translations in the same way, that is, multiplying by a matrix (i.e., a $4 \times 4$ matrix).
Remember that without these $4 \times 4$ matrices, translations would be represented by summing with a vector, whereas rotations and scaling are represented using multiplications with respectively a vector and a scalar factor.
Now the question is: how do we pass from a 3D coordinate systems to a 4D one? The answer is "homogenous coordinates".
So, what does it mean? We construct $4 \times 4$ matrices to represent rotations, scaling and translation, so that we only use matrix multiplications to represent transformations (e.g., rotations, scaling, etc). How we construct them individually, it's more specific, but you can have a look it on the web.
At this point, we have $4 \times 4$ matrices and 3D vectors, not useful yet, because you can't multiply $4 \times 4$ matrices and $3D$ vectors, since the dimensions do not match. That's why, when we work with homegeneous coordinates, we also need to convert our given 3D points into corresponding 4D ones.
How do we do it?
We distinguish between direction and position vectors. Direction vectors, as the name suggests, have a direction at which they are pointing; we also care about their length, but they are not affected by translations, since we don't care about their position. Position vectors (or simply "points") can be translated or moved around; they are usually represented with respect to the origin, i.e. as a vector from the origin to the point itself.
We transform 3D direction vectors by adding a $0$ as the $4$th coordinate of the corresponding homogeneous vector: we add a zero because this basically eliminates the effect of translations. We do a similar thing with position vectors, but instead of a $0$ we add a $1$, for the opposite reason.
For example, if we have a $3D$ direction vector $v = \begin{pmatrix} v_1 \\ v_2 \\ v_3\end{pmatrix}$, we transform it by doing $v' = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ 0\end{pmatrix}$. Similarly, if we had a point vector $u = \begin{pmatrix} u_1 \\ u_2 \\ u_3\end{pmatrix}$ we would transform it to $u' = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \\ 1\end{pmatrix}$
Note: to convert from homogeneous coordinates back to corresponding $3D$ ones, you can't simply remove the $4th$ coordinate, unless it's still equal to $1$ (or $0$ respectively).