what does project away mean? I realise I should know this but I have no idea what people mean when they say "we project away from this point" (or replace point with line, plane or whatever in projective space). What does this mean?
 A: If $L\subset \mathbb P^n$ is a linear subspace of dimension $l$ and $M\subset \mathbb P^n$ is a linear subspace of dimension $m=n-l-1$ such that $L\cap M=\emptyset$, the projection from $L$ onto $M$ is the morphism $\mathbb P^n\setminus L\to M$ sending $p$ to the unique point of intersection $M\cap \overline {pL}$ of $M$ with the linear subspace generated by $p$ and $L$.  
This is quite close to the way Renaissance painters invented projective geometry: they projected the three-dimensional world from their eye, the zero-dimensional $L$, onto their two-dimensional canvas $M$.
How not to be nostalgic about  a time when projective geometry meant the art of  the likes of Della Francesca, Brunelleschi, Donatello or Alberti?
[Please, excuse the lyric tone: I'm just back from a holiday  in Florence ...]  
Edit: a concrete computation
As an attempt to answer carmelo's question in his comment, let me make an explicit computation. 
Consider $l=[0:0:1]\in \mathbb P^2$ (with coordinates $x:y:z$) and let us project onto the line $M$ of equation $z=0$ .
We get the map $f:\mathbb P^2\setminus \{l\}\to M:[x:y:z]\mapsto [x:y:0]$, a rational map from $\mathbb P^2$ to $M$ which cannot be extended regularly through $l$.  
However if we blow up $l$ we get the variety $\tilde {\mathbb P^2}\subset \mathbb P^2\times \mathbb P^1$ defined by the condition $([x:y:z][u:v])\in \tilde {\mathbb P^2}$ iff $xv=yu$ (where $\mathbb P^1$ has coordinates $u:v$).
The rational map $f$ now extends to the regular morphism  $$\tilde f:\tilde {\mathbb P^2} \to M:([x:y:z],[u:v])\mapsto [u:v:0]$$
The indeterminacy of $f$ at $l$ has been resolved by blowing up the point  $l$ to the copy of $\mathbb P^1$ formed  by  all the pairs $([0:0:1],[u:v])\in \tilde {\mathbb P^2}$
