No it won't. It'll transfer charge until both batteries are at the same potential. If the batteries are not under load at the time, this is not likely to be too much of a problem. The two batteries will, between them, contain the same energy (minus a tiny bit during the transfer). A problem would arise if one battery were damaged and was not able to reach the potential of the other.
However, since a switch is being fitted, I presume it will be used. Putting the switch onto the 'both' setting can be done before starting the engine and, once the engine is started, both batteries will be receiving a charge. If, when leaving the boat, the switch is moved to the off position, there won't be a problem, even with a damaged battery. Having said that, what is the point of keeping a damaged battery on the system?
Richard's point about not jumping the positives for the purpose of supplying the bilge pump, is a good one as is Chris's point about drawing a diagram and looking for errors before you get near to the actual wiring up
My own preference is to use a switch but, also, to add another in the form of the screw type battery isolator which is commonly fitted to cars as an accessory. These fit into the negative battery post. I use one battery as the main supply, fit the isolator to the other, and I bridge the isolator with a schottky diode. The isolator is then left open. The switch is set to 'both' during use and both batteries will charge.
Whether the engine is running or not, the main battery is the one in use because the diode blocks the second one from discharging. Should the first battery become discharged or damaged, closing the isolator will give a fully charged battery, ready for service.
If it's a discharge situation, once the engine is running, opening the isolator will allow the the system to revert to normal and the discharged battery will recharge as normal.
The use of a schottky diode limits the voltage drop for charging and ensures the backup battery reaches a good state of charge. The normal limits for the voltage regulator are 13.7v to 14.2v so the 0.4v loss in the diode is more than tolerable. Unless you've got a lowish output from the regulator,
then the whole system will be a bit down.
These things always sound more complicated by description than they actually are.
Go slowly, take care to search out any errors and it'll be fine.