First, I'm going to call the 3 piles to be deciphered Deck 3, Deck 2, Deck 1, to correspond with the number of deciphered cards required for the "phrase."

I like it. I'll change it to read this way, too.

As I understand it, we slide the top card (I'll call it the "slider") down letter-by-letter until the hidden card's title is completely revealed. That revealed ("deciphered) title card is then placed beside that deck until the required number of cards for a "phrase" (3, 2, or 1)

The "slider" card is placed aside. This makes the next card that you just 'revealed' to be the new slider, and so on. This makes the motion of slide/revealing always a discovery/mystery, and very fluid if you're working on a long phrase.

When the phrase is fully deciphered, the player who completed it takes the slider card from that deck into his discard pile (to be shuffled and drawn as needed for our 3-card hand to begin each turn.)

When the phrase is fully deciphered, the play takes all previous slider cards that have been set aside. In the case of the three-card phrase it would be all three slider cards. In the case of "Deck 1" (pawn, court, and The Excuse Deck) you would just collect the slider card each time you successfully decipher a one-card phrase.

If a phrase is partly deciphered, say only 2 of the 3 required cards are beside Deck 3, it is possible that the opponent may be able to claim the cards by finishing the reveal of the 3rd. Is this correct? So player A might have revealed 3 of the four letters of the card's title, for example, but he has no cards left in his hand and his turn is over. Now Player B, on his turn, could reveal the final letter, and thus claim the phrase card(s) next to the deck. Is this correct?

That is all correct.

I don't quite understand that "overshoot the phrase" business. If we play a 3, for example, but there are only 2 unrevealed letters on a card we're working on, what card(s) does the opponent get? Just the card that was overshot, or that card plus any other phrase cards which are next to that deck?

If you 'overshoot the phrase' by any amount, all the cards that were set aside for that phrase go to your opponent's discard. For the case of Deck 1, it would just be the current slider card. For the case of Deck 3 it would be all three slider cards.

Over on the BGG forum you suggested examples of play. I'll add them here.

Thanks again.

First, I'm going to call the 3 piles to be deciphered Deck 3, Deck 2, Deck 1, to correspond with the number of deciphered cards required for the "phrase."

As I understand it, we slide the top card (I'll call it the "slider") down letter-by-letter until the hidden card's title is completely revealed. That revealed ("deciphered) title card is then placed beside that deck until the required number of cards for a "phrase" (3, 2, or 1), When the phrase is fully deciphered, the player who completed it takes the slider card from that deck into his discard pile (to be shuffled and drawn as needed for our 3-card hand to begin each turn.)

If a phrase is partly deciphered, say only 2 of the 3 required cards are beside Deck 3, it is possible that the opponent may be able to claim the cards by finishing the reveal of the 3rd. Is this correct?

So player A might have revealed 3 of the four letters of the card's title, for example, but he has no cards left in his hand and his turn is over. Now Player B, on his turn, could reveal the final letter, and thus claim the phrase card(s) next to the deck. Is this correct?

I don't quite understand that "overshoot the phrase" business. If we play a 3, for example, but there are only 2 unrevealed letters on a card we're working on, what card(s) does the opponent get? Just the card that was overshot, or that card plus any other phrase cards which are next to that deck?

Sorry if I'm a little thick. I blame it on my age.