By that statement, does that mean adding a random character to a random position in a Diceware word adds 10 bits to each word?
No. The ten bit estimate is for adding a random symbol from the 36-item table to a random position in the passphrase. The entropy in the character choice is about five bits and the entropy in the choice of position is another almost five. The exact number depends on passphrase length, but six word passphrases are ~30 characters long on average.
If you add a symbol at random to each word, you lose more than half the "position" entropy, and get ~7 bits per added symbol.
(The entropy loss from duplicates that Nova mentions in the other answer is minimal, because most diceware words are primarily letters, while the added characters are digits or symbols.)
What would be the entropy per Diceware word if random printable ASCII character is used instead?
The entropy in a printable ASCII character is about 6.5 bits, so you could only gain 1.5 bits if you did that (so about 11.5 bits in total). Here there would be a loss from duplicates, as well as a more serious loss from being able to accidentally get another Diceware word (e.g. woo + d -> wood). Further, it would be more difficult to generate when using dice.
What would be the entropy per Diceware word if a random Diceware word is used as the random symbol in a random position per Diceware word?
Well, the entropy in a Diceware word is 12.9 bits, so you'd gain ~18 from adding one randomly anywhere in the passphrase or ~15 per from adding one to each word. However, that could again form another Diceware passphrase (e.g. yam + aha -> yamaha), so the actual entropy added would be slightly less.
Note: This problem also exists when creating a normal Diceware passphrase, which is why they recommend always using a space as a separator between words.