Those are bitwise math operators.  Check wikipedia/google for a more thorough explanation than what I'll give.

So "<<" is left shift.  Which means bank is shifted 14 bits to the left.  This usually means bank * 2^14 (any bits that move "past" the register's max size are cut off)

& is bitwise AND.  $3FFF is 0011 1111 1111 1111 in binary.  You convert pointer to a value like so (say the pointer is $63F8) and do the following operation (only when both bits is 1, the result is 1...otherwise result is 0):
0011 1111 1111 1111 &
0110 0011 1111 1000 =
0010 0011 1111 1000 which is $23F8

& is basically a way to zero certain bits you don't want.

Windows Calculator can do these operations for you.  In Win7, change to Programmer mode.  (Vista might be the same...XP it was in scientific mode)

PS: Next time stick around in IRC for longer than a minute to give people a chance to answer your question.  The room may have many people, but most are idle.

0x0D shifted left 14 bits is 0x34000. And the pointer is 0x4000. You tried reading the pointer in "big-endian" order instead of "little-endian" order.

I think since you were in hexadecimal mode, the "14" was interpreted by the calculator as 0x14 (20 in decimal) instead of a decimal 14 (0x0E in hexadecimal).

Both are bitwise operators. They are an essential notion in ROM Hacking and programming in general. You use them to manipulate binary values in various ways. You should get familiar with these operators as a first step in ROM Hacking, along with hexadecimal notation. Wikipedia has a good page about them: http://en.wikipedia.org/wiki/Bitwise_operation.

The << means shift bits to the left. Mathematically speaking, it effectively multiplies your value by 2.
The & means bitwise AND. Used to keep (or "mask") specific bits. Not to be confused with &&, which is the logical AND, used in boolean logic.

In your specific case, the value of "bank" is shifted 14 times to the left, then added to the value of "pointer" with the 2 most significant bits masked (set to 0).
Let's say "bank" = 3, and "pointer" = ABCD.

0003 in binary -> 0000 0000 0000 0011
ABCD in binary -> 1010 1011 1100 1101
3FFF in binary -> 0011 1111 1111 1111

0003 <<  14 = 1100 0000 0000 0000 = C000
ABCD & 3FFF = 0010 1011 1100 1101 = 2BCD
C000 + 2BCD = EBCD

So your final pointer value is EBCD. You may have to add an arbitrary value (like 0x10) to account for the ROM header, if there is one...