As a set, the vector space $k^n$ and the affine space $k^n$ are the same but as spaces they are thought as having slightly different structures over them. As a vector space you have a neutral element $\vec{0}$ under addition and two vectors can be added together $\vec{v}+\vec{u}$, however as an affine space you think of the elements as points generalizing Euclidean space: you do not distinguish a particular origin $\vec{0}$, you cannot add points but you can attach the vector space $k^n$ to a particular point $P$ of the affine $k^n$ establishing a reference system and allowing you to substract points considering the vector from one to the other $P-Q:=\vec{PQ}$; thus you can reach any other point by the action of any vector onto your chosen point of origin $Q=P+\vec{v}$. Therefore an affine space $A$ is a set of points together with a reference system (think of coordinates) given by the action of a vector space $V$ on it, so $(A=k^n, V=k^n)=:\mathbb{A}^n_{k}$ is a bigger structure than $V=k^n$ by itself.
The affine space $\mathbb{A}^n_{k}$ is not a projective space because for example it is not compact whereas any projective space is a compact topological space. Moreover, a projective space $\mathbb{P}^n_{k}$ is constructed in a different manner than affine space: given a reference frame (origin) on $\mathbb{A}^n_{k}$, think of all the straight lines that go through it and parametrize them by a set. In this sense, whereas the points of $\mathbb{A}^n_{k}$ are in one-to-one correspondence with $k^n$, the points of $\mathbb{P}^n_{k}$ are in one-to-one correspondence with one-dimensional vector subspaces of $k^{n+1}$, i.e. a point of $\mathbb{P}^n_{k}$ is an equivalence class of points $(x_1,...,x_{n+1})$ where any other $(\lambda x_1,...,\lambda x_{n+1})$ is in the same class for all nonzero $\lambda\in k$. You should think of this as $\mathbb{A}^n_{k}$ with an infinite point added for every direction. For example the real projective plane $\mathbb{P}^2_{\mathbb{R}}$ can be showed to be $\mathbb{A}^2_{\mathbb{R}}\cup \mathbb{P}^1_{\mathbb{R}}$ which is a compactification of the affine plane since you are adding a circunference boundary at infinity with every point identified with its opposite (think of a circe, the interior would be your affine plane and the circunference would be $\mathbb{P}^1_{\mathbb{R}}$ where if you travel from the origin and reach the boundary you reappear in the opposite point of the circunference). Nevertheless projective spaces can be charted by affine spaces, like manifolds can be charted by $\mathbb{R}^n$, since they are the easiest example of projective varieties.
Projective varieties are introduced because for example, working with compact spaces is much better technically (e.g. you can do integration over the whole space). In algebraic geometry working projectively is natural, since adding the points at infinity simplifies and unifies a lot of results: for example Bézout's theorem is a great simple result which needs the possibility of two curves intersecting at infinity. This is the case of two parallel straight lines which do not intersect in $\mathbb{A}^2_{k}$ but can be thought to meet at the same point at infinity, i.e. intersect within $\mathbb{P}^2_{k}$. The name "projective" actually reflects this since its origin comes from descriptive geometry where two parallel lines in perspective meet at a single point of the horizon (line of sight).
There are similar reasons to work over the complex numbers $k=\mathbb{C}$ or any other algebraically closed field so that your polynomial equations have always solutions. That is the reason why most of the time one wants to work in complex projective space $\mathbb{CP}^n$ or in its projective subvarieties.
Thus, the advantages of projective varieties are many and fundamental to the development of geometry.
A projective bundle $\mathbb{P}(\mathcal{E})$ over a smooth projective variety $X$ (over any base field $k$) is indeed a smooth projective variety.
Such a scheme $X$ is noetherian. By Exercise II.7.10. in Hartshorne, for a locally free sheaf of rank $n+1$ on $X$, its projectivization $\mathbb{P}(\mathcal{E})$ is always a $\mathbb{P}^{n}$-bundle over $X$, and conversely, since $X$ is also regular, every $\mathbb{P}^{n}$-bundle over $X$ arises in this way.
By definition (see here) the morphism $\pi \colon \mathbb{P}(\mathcal{E})\to X$ is projective in the sense of EGA. But $X$ admits an ample invertible sheaf, so in this case EGA-projective imlpies Hartshorne-projective (see the same reference a bit below). Since composition of Hartshorne-projective morphisms is Hartshorne-projective and the structure morphism $X\to \text{Spec}(k)$ is Hartshorne-projective, so is $\mathbb{P}(\mathcal{E})\to \text{Spec}(k)$ and therefore $\mathbb{P}(\mathcal{E})$ is a closed subset of some projective space over your base field $k$.
Smoothness of $\mathbb{P}(\mathcal{E})$ over $k$ follows as you say from the trivializations: smoothness is a local property and $\mathbb{P}(\mathcal{E})$ is locally a product of an open set of $X$ (smooth) and a projective $n$-space over $k$ (also smooth).
Irreducibility can be shown as follows: if $U$ is a trivializing open in $X$, by irreducibility of $X$, $U$ is also irreducible. Now $\pi^{-1}(U)$ is the product of two (irreducible) quasi-projective varieties over $k$, hence also irreducible. But in fact $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$, so we get that $\mathbb{P}(\mathcal{E})$ is irreducible.
To see that $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$ you have two arguments:
The (topological) map $\pi$ is open and therefore the preimage of a dense subspace is dense.
For any other trivialising open $V$, the intersection $U\cap V$ is dense (again by irreducilibity of $X$) and so the preimage $\pi^{-1}(U\cap V)=\pi^{-1}(U)\cap \pi^{-1}(V)$ is dense in $\pi^{-1}(V)$. But these sets cover $\mathbb{P}(\mathcal{E})$, hence $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$.
Best Answer
Let's consider an explicit example. Look at the equation $xy=z^2$ in the projective plane $\mathbb{P}^2$ with coordinates $[x:y:z]$. The given locus is a quadric, i.e. a curve isomorphic to $\mathbb{P}^1$ and with the property that its intersection with a line (a copy of $\mathbb{P}^1$ given by linear equations) is two points (including the case of one point with mutliplicity 2).
Now, the way we build the cone is the following. Remember that $\mathbb{P}^2$ with coordinates $[x:y:z]$ is obtained from $\mathbb{A}^3$ with coordinates $(x,y,z)$, removing $(0,0,0)$, and quotienting by the rescaling action of the group of units of the ground field $k^*$. In particular, the equations that define our quadric (or, more generally, the projective variety in $\mathbb{P}^n$ you start with) still make sense in $\mathbb{A}^3$ ($\mathbb{A}^{n+1}$ respectively). Some algebra computations show that the locus you obtain has one extra dimension than what you started with. The reason is, as you correctly wrote, that we are taking the space of lines over the projective variety.
It is important to stress that we are not considering these lines as points in the projective space, but as honest lines in affine space. Thus, the picture that the real points (i.e. the points that live over $\mathbb{R}$) of the above example are the following: you can think of the projective conic as a cricle, and the cone over it is the honest right cone with circular base.
Thus, the affine cone over a projective variety is a cone whose ''horizontal slices'' recover the variety you started with. Notice that the cone is always singular at the origin.
By construction, the geometry and the properties of these two objects are closely related. For instance, the cone is a normal variety if and only if the embedding of the projective variety is projectively normal.
Cones are also very important, since they provide a nice list of examples of computable singular varieties, where one can test ideas and computations about singular varieties.