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For many questions, the easiest way to see the nuts and bolts of a projective variety $V \subseteq P^n$ is to look at its cone $CV \subseteq A^{n+1}$. After all, the graded ring whose Proj is $V$ is the same as the ungraded ring whose Spec is $CV$. Obviously, there is almost always a singularity at the origin; but if you ignore that point, the other singular points all correspond between $V$ and $CV$. You can also think of the grading as geometrically represented by multiplication by $k^*$, if you are working over an algebraically closed field $k$. (Because the homogeneous polynomials are then eigenvectors of that group action.) You can think of $V$ as obtained from $CV$ by and then dividing by scalar multiplication.

The atlas-of-charts analysis of a projective variety is certainly important, but to some extent it is meant as an introduction to intrinsic algebraic geometry rather than as the best computational tool.

Your second question is reviewed in Wikipedia. As Wikipedia explains, Hironaka's big theorem was that it is possible to resolve all singularities of a variety by iterated blowups along subvarieties. I do not know a lot about this theory, but if so many capable mathematicians went to so much trouble to find a method, then surely there is no simple method.

On the other hand for curves, there is a stunning method that I learned about (or maybe relearned) just recently. Again according to Wikipedia, taking the integral closure of the coordinate ring of an affine curve, or the graded coordinate ring of a projective curve, solves everything. The claim is that it always removes the singularities of codimension 1, which are the only kind that a curve has.

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For many questions, the easiest way to see the nuts and bolts of a projective variety $V \subseteq P^n$ is to look at its cone $CV \subseteq A^{n+1}$. After all, the graded ring whose Proj is $V$ is the same as the ungraded ring whose Spec is $CV$. Obviously, there is almost always a singularity at the origin; but if you ignore that point, the other singular points all correspond between $V$ and $CV$. You can also think of the grading as geometrically represented by multiplication by $k^*$, if you are working over an algebraically closed field $k$. (Because the homogeneous polynomials are then eigenvectors of that group action.) You can think of $V$ as obtained from $CV$ by and then dividing by scalar multiplication.
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