Should I use step-up or step-down transformer?
Since you mentioned transformers and \$12\$V adapters, I assume you are stuck in the old ways of linear power supplies, which typically only have a power efficiency of between \$50\$%-\$60\$%

So if you go from \$120\$V AC at the wall to \$12\$VDC to \$5\$VDC there is a \$50\$% worst case power conversion loss at each stage.  It won't matter if you step up or step down, you can assume that the up-conversion or down-conversion is equally efficient, and the losses are more from the linear regulators than anything else. 

In this day and age, you should really only be using switching power supplies, especially for digital electronics like a tablet. There are numerous designs available and numerous chipsets with efficiencies approaching \$98\$% these days. 

If you go the switching regulator route, then the path will be \$120\$VAC -> \$12\$VDC -> \$5\$VDC and you can expect \$80\$%-\$90\$% efficiencies at each stage. This is the path I would use if you were coming from the wall, because this is two buck regulators in series and a buck regulator is inherently more efficient than a boost regulator. This is because a buck regulator is either on or off, and in the on cycle it is delivering current to the load, and in the off cycle the inductor is doing it. However, a boost regulator first must deliver current to ground through the inductor to build up the inductor's magnetic field, then in the off cycle (and during the collapsing field) the inductor delivers current to the load. So the resistive losses of the inductor during the buildup phase of a boost regulator are wasteful (however, modern boost controllers go through extra hoops to minimize this wasted energy)



However, if you are working off a battery source (you did say "tablet", so I assume portable), then your choice of buck vs boost is often directed by your source voltage available. If your source is a couple of lithium-polymer cells, then you only have a few volts to start with, and so you will need a battery -> boost \$5\$V -> boost \$12\$V configuration.