Just to be sure, MAGOO is the AdS/CFT Bible
http://arxiv.org/abs/hep-th/9905111
whose authors are usually listed alphabetically. Concerning your questions,
Conformal dimension or scaling dimension or mass dimension are meant to be the same thing in the context of conformal field theories in more than 2 dimensions. In 2 dimensions, one may distinguish the left-moving (holomorphic) and right-moving (antiholomorphic) dimensions or their sum - but the separate chiral "dimensions" are usually called "weights", anyway. The dimension pretty much counts the exponent of "kilogram" in the unit of the corresponding operator - except that in quantum mechanics, the exponent is often fractional. The dimensions are usually positive - they get mapped to the energy which is positive as well. Operators have positive dimensions of mass. If your cryptic "-ve" meant that there is a minus sign, then there is no minus sign. If you prefer to use units of length, then their powers are negative, but you should switch to masses as the base whose powers count the dimension.
a) The field $\phi$, as opposed to $\phi_0$, is dimensionless because it is an actual field in the AdS space (the bulk), unlike $\phi_0$ that is defined on the boundary. They mean that it is dimensionless under the conformal transformations of the boundary - and that's true because the conformal transformations are realized as simple isometries in the bulk, so they can't rescale the AdS field $\phi$ - at most, they move it to another point. Also, on that page, one deals with particular finite modified boundary conditions for $\phi$ at the boundary of the AdS space so it must be finite even at $z=0$.
b) In equation (3.15), the left hand side is dimensionless. The right hand side has a power of $\epsilon$ whose dimension is $length^{d-\Delta}$ because $\epsilon$ has units of length (on boundary), and $\phi_0$ whose dimension must be $length^{\Delta-d}$ as a consequence, to get a dimensionless product. In equation (3.13), the exponent on the left hand side has $d^4x$ whose dimension is $length^d$ - note that $d=4$ for $AdS_5$; $\phi_0$ whose dimension is $length^{\Delta-d}$ as I said in the previous sentence - note that the $d$ term in the exponent cancels; and $O$ whose dimension must therefore be $length^{-\Delta}$ which means $mass^{\Delta}$. As I said, the standard base for counting dimensions is mass, so we also say that $O$ has dimension $\Delta$. So I suspect that your sign error is simply caused by the point 1) - namely by your incorrect assumption that dimensions refer to the powers of length.
When we say "dimension" without extra specifications, we mean the power of the mass, not length. It's a healthy convention because the operators end up having non-negative dimensions. You may want to remember the dimensions of basic operators in 4 dimensions. The identity operator is always dimensionless (and $x$-independent): the dimension is 0. Bosonic scalar fields $\phi_0$ and potentials $A_\mu$ have dimension 1, their field strength has dimension $2$, fermions like the Dirac $\psi$ have dimension 3/2, and all terms in the Lagrangian density have dimension $4$. These are classical dimensions; each derivative adds $1$ to the dimension. Products of these operators have dimensions that are sums of the dimensions of the factors - except that quantum mechanics also adds "anomalous dimensions" to this classical form of the dimension, so that the total dimensions may have fractional terms that are proportional to powers of $g$ etc.
