The first line simply sets all the bits of usVar to 0.
The second line is equivalent to
usVar = usVar | (1 << 14); // parentheses because I don't remember operator precedence
is the "left shift" operator. It moves the bits representing a value by a number of places specified by the number on the right.
For example let's do 7 << 2
7 = 00000111
Moving the bits to the left by 2 places yields 00 | 00011100. The leftmost bits are lost, and the hole is filled with 0s.
Now, 1 = 00000001. You can see that only the first bit is on. Left-shifting by N means that you will turn on only one bit, the one N positions away from the rightmost bit, and the first bit will be left with a 0. This allows you to turn on a specific bit, and to turn on that bit for another value you use the bitwise OR.
is the bitwise OR operator. You need to know some boolean algebra here; if you do you can easily understand that
usVar = 00000000 OR
1<<5 = 00100000
you turned on the 5th bit of usVar. (counting from 0, or 6th from 1)
I used 8 bits for simplicity.
Important note: bit shifting is relatively easy with unsigned types, which use logical shift. For signed types it's more complex, they use arithmetic shift.