What youre describing is a help mate. Your diagrammed position is not a mate with correct play.
Draw by insufficient material rule

By FIDE rules it has to be a forced mate.
Well... if one side runs out of time with all the pieces still on the board (black and white), then the other side doesn't have a forced mate, but the game is declared a win for the side that did not timeout. So where is the line drawn? Is is based on the number of pieces on the board - if it's fewer than x, and one side runs out of time and the other doesn't have a forced mate, the game is drawn? And if it's greater than x, then whoever runs out of time loses?
By FIDE rules it has to be a forced mate.
That's incorrect. FIDE simply requires checkmate to be possible from the position on the board at the end of the game. It does not have to be forced. "The game is drawn when a position has arisen in which neither player can checkmate the opponent’s king with any series of legal moves."
According to USCF rules with a lone king, king and bishop, king and knight, or king and two knights, and with no forced win the game is drawn.
The Chess.com insufficient material rule is based on the USCF rule but it does not consider whether a forced mate is possible.

Rule states that. If a player suppose white has 2 knight and opponent has only king a draw can be claimed as 2 knights can not force a mate.
And if you have 1 rook and opponent has only king and you run out of time it is a draw because your opponent cannot mate you

So Chess.com need to declare whose rules govern? USCF or FIDE? If the latter, NewEngland has a point ( eg K and N v K and P, depending where the pawn is). If the former, win must be forced.
In most cases with basic opposition knowledge king pawn endgame can be drawn. Note:- If a player with only king left on board to play has opposition it is a draw. But this draw cannot be claimed.

It seems to me that the "forced mate" interpretation is the easiest to apply, and would be understood by everyone. Surely a footnote in the Chess.com rules to this effect would clarify the position once and for all?
That is right simplest way to understand this is either of side cant force mate other it is a draw.
I one side has material to force mate the opponent but opponent doesn't have this material and if the side having the material to mate runs out of time it is a draw
"And if you have 1 rook and opponent has only king and you run out of time it is a draw because your opponent cannot mate you "
That's completely wrong, of course. You can always force mate with K+R against K
But not with K against K+R.
The USCF rule seems more useful, as not even a novice would manage lose KNKN or KNNK. The same could be said for KRKR in positions where there isn't a forced win, however.

I have to also ask this question as it has happened to me once where I have two rooks and a queen and a king vs. a king and a pawn--I ran out of time before I mated my opponent and he was credited with a win. So if we are talking about forced mates, I must ask, in the following position, if black ran out of time before mating white, should this be a win for white or a draw? These positions are often credited as a win for the colour that has the mere pawn despite the fact no forced mate exists.

"Rule states that. If a player suppose white has 2 knight and opponent has only king a draw can be claimed as 2 knights can not force a mate."
I don't think you can claim a draw when your opponent can possibly checkmate you still with his remaining pieces--best avoid blundering.

I have to also ask this question as it has happened to me once where I have two rooks and a queen and a king vs. a king and a pawn--I ran out of time before I mated my opponent and he was credited with a win. So if we are talking about forced mates, I must ask, in the following position, if black ran out of time before mating white, should this be a win for white or a draw? These positions are often credited as a win for the colour that has the mere pawn despite the fact no forced mate exists.
Opponent had a pawn and he could promote a queen and then force a mate even if your opponent has a pawn and you run out of time having material to mate then you lose

"Opponent had a pawn and he could promote a queen and then force a mate even if your opponent has a pawn and you run out of time having material to mate then you lose"
And so then why is a position where there is king and knight vs king and knight and one side runs out of time a draw? Both require an incredible blunder-fest on one side for a mating possibility for either side to show up but a possibility to mate exists. And no, there is no forced mate that white has. All black has to do is merely capture the pawn with one of his pieces and he has two rooks or a rook and queen against the lone king. So no mate actually exists for white unless a blunder-fest is presumed. Likewise, in situation of K+N vs. K+N no mate exists unless there is a blunder-fest. So if a blunder-fest is allowed to presume the final position at the end of the game (take another game I won on time last night where I had no forced mate unless a blunder-fest occurred), then how is it different for K+N and K+N?

I need to ask another question--how are we defining a forced mate? It seems the position I have posted where black has two rooks, a pawn, a queen, and a king vs. white's mere king and pawn has no forced mate yet some seem to argue that since the pawn can promote to a queen and black can blunder all of his pieces away, a forced mate position can arise.
Are we defining a forced mate position as a possibility to arise or are we defining a forced mate position as mate in x number of moves for one colour? If the primer, I must ask--can the following position with black to move show up in my own game recently?
If the answer is "yes" is this not a forced mate in one for black?

Wantobegm,
Is the example I have provided where black has two rooks, a queen, a pawn, and a king, vs. white's king and pawn a forced mate? Can a forced position ever arise in such situation where there exists a forced mating possibility?
If the answer to one, the other, or both, is "no" then where does the difference come in when describing the situation of K+N vs. K+N? (The same questions can also apply to the K+N vs. K+N position.)
I asked a major question as to how we are defining "forced mate". Please no dodging the questions at hand.

it is possible to promote the pawn and get a queen then it is forced mate
But you cant force mate with 2 knights

"Lets not complicate it. Lets say you are white with K and R, and black's time runs out. Surely the essence of the rule is that white has material enough for a forced mate, even though black might have had all his pieces when his time ran out. White wins, irrespective of black's strength."
If we define "forced mate" then as "having sufficient material to produce a forced mate" or "having enough material for a forced mate" then perhaps this is a valid argument. It still complicates things as it must be asked what then is the intended meaning of the rule.
And surely, even situations where a K+N vs. K+N can lead to a situation (though unforced) where one side can attain a forced mate?

"it is possible to promote the pawn and get a queen then it is forced mate"
It is only possible provided a blunder-fest. Likewise, in issue of K+N vs. K+N it is possible for it to be black to move with his knight on e5 and white's king on h8 and white's knight on g8 and the white king on g6. Then it is forced mate in 1 for black. Both require blunder-fests but both situations are possible.

So Chess.com need to declare whose rules govern? USCF or FIDE? If the latter, NewEngland has a point ( eg K and N v K and P, depending where the pawn is). If the former, win must be forced.
Actually in this instance Chess.com rules govern To keep things simple, our insufficient material rule only considers material. There are no edge cases wherein forced mate could conceivably happen from the final position. The only thing that matters is what material you have left when your opponent's time runs out.
I'm wondering if someone could explain to me the full rule here on insufficient material draws when the opponent runs out of time but you have insufficient material.
Some situations, it seems that in full theory, if one has a piece that technically cannot deliver a forced mate, a mate is still possible.
So for instance, this game:
And now, a presentation of a continuation (my opponent ran out of time and the game was drawn by insufficient material) where black can mate white.
I hear the explanation a lot where when the one guy has just a king that the one guy cannot mate the other guy but in full technicality, both sides have mating chances here so I'm not certain why the draw by insufficient material. Given, it's not a forced mate, but unless there is a nuance in the rule I am not aware of, I'm confused why situations like these are draws.
Could someone explain the entire rule please?