Let $X$ be an affine variety and let $X'$ be its embedding to projective space (by embedding affine space to $x_0 = 0$). Then given a regular function on $X$ (say $f$) you get a regular function on $X'$ by homogenizing $f$ and then dividing by $x_0$ to the degree of $f$. Conversely given a regular function $g\over h$ on $X'$ you get a regular function on $X$ by setting $x_0=1$. This correspondence gives the isomorphism between ring of regular functions of $X$ and that of $X'$.
Since we embed isomorphically, it doesn't matter how we embed (you can find an isomorphism between two embedding by composing the inverse of one embedding with the other embedding).
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.
Best Answer
Yes, you can use the affine cone to check singularities. The key step is to prove the projective Jacobian criterion:
Projective Jacobian criterion: Let $Y\subset\Bbb P^n$ be a projective variety of dimension $r$ with homogeneous ideal generated by $f_1,\dots,f_t\in S=k[x_0,\cdots,x_n]$. If $P\in Y$ is a point with homogeneous coordinates $(a_0,\dots,a_n)$, $Y$ is nonsingular at $P$ if and only if the rank of the matrix $|\frac{\partial f_i}{\partial x_j}(a_0,\dots,a_n)|$ is $n-r$.
With this result, it's immediate that the affine cone on $Y$ (the variety in $\Bbb A^{n+1}$ with coordinates $x_0,\dots,x_n$ cut out by $f_1,\dots,f_t$) is nonsingular away from the vertex if and only if the projective variety is nonsingular, since the projective Jacobian criterion is exactly the usual affine Jacobian criterion for the cone away from the vertex.
I think it's probably better if you prove the projective Jacobian criterion for yourself than if I do it for you. Here's an outline to follow:
Feel free to leave a comment if you run in to trouble.