ag.algebraic geometry - Why is the fibered coproduct of affine schemes not affine?

The short answer is that the category of affine schemes does have pushouts, but these are not the same as pushouts of affine schemes calculated in the category of all schemes.

For a longer answer, consider an example that's small enough to compute: The projective line (over the complex numbers, say) is not affine but it is gotten by gluing two copies of the affine line along the punctured affine line, so it is the pushout (in the category of schemes) of a diagram of affine schemes. Now what's the pushout of that same diagram in the category of affine schemes? Well, the rings involved are two copies of $C[x]$ and a copy of $C[x,x^{-1}]$. The two maps are the two injections of $C[x]$ into $C[x,x^{-1}]$, one sending $x$ to $x$, and the other sending $x$ to $x^{-1}$. The pullback of these, in the category of commutative rings, is just $C$, because the only way a polynomial in $x$ can equal a polynomial in $x^{-1}$ is for both of them to be constant. Therefore, the affine-scheme pushout is not the projective line but a point.

Intuitively, if you glue together two copies of the line along the punctured line "gently," allowing the result to be non-affine, you get the projective line, but if you demand that the result be affine then your projective line is forced to collapse to a point.