If I've understood correctly, the tool we have is not a balance scale as in the traditional 12-coins puzzle but a weighing scale that tells you (say) how many grammes the coins you put on it weigh in total. The question doesn't make clear whether we're supposed to know the "standard" weight of the coins. I'll address both the version of the question where we do know this, and the version where we don't.
If we do know the nominal weight of the coins
Given that only one coin is odd, and that we know how many coins we're weighing on any given occasion, what the scale tells us is exactly the following: Is the odd coin on the scale or not, and if so is it heavy or light?
each such weighing gives us at most three different results, and after the first (if any) whose result isn't "odd coin not on scale" the heavy-or-light information tells us nothing new. I think a counting argument then indicates that we must take at least four weighings.
label the coins A..L and keep them a consistent way up throughout. First weigh ABCDEF. If the odd coin is not among their number, we now have GHIJKL to assess in three weighings. Now weigh GHIJ. If the odd coin isn't one of these then it's either K or L; weigh them separately. Done in four weighings. On the other hand, if weighing GHIJ indicates that the odd coin is one of them, it also tells us whether it's heavy or light. Weigh GH, and then either G or I depending on the result. Done, again, in four weighings.
On the other hand,
if the odd coin is one of ABCDEF then we can do to them what we did above to GHIJKL (we don't even need to use our knowledge of whether the odd coin is heavy or light). Done, again, in four weighings.
If we don't know the nominal weight of the coins
Still keeping all coins a consistent way up unless stated otherwise, first weigh ABCD and EFGH (two weighings). If they come out the same then the odd coin is one of IJKL and we now know the nominal weight of a coin. Weigh IJK. If it comes out right then L is the odd coin and one more weighing will tell us which way it's wrong. If not then we know, say, that one of IJK is too heavy. Now weigh I plus J-upside-down. If this is heavy then I is odd; if light then J; if neither then K. Done in four weighings.
On the other hand,
if ABCD and EFGH gave different results then call them $x$ and $y$ where, let's say, $x>y$. Then write $\delta=x-y$; we know that either normal coins weigh $y/4$ and one of ABCD weighs $y/4+\delta$, or else normal coins weigh $x/4$ and one of EFGH weighs $x/4-delta$. Turn EFGH the other way up, so now the odd coin definitely weighs $\delta$ "too much", and weigh ABCEFG. If the odd coin is in ABC, this will weigh $6y/4+\delta$; if in EFG, it will weigh $6x/4+\delta$; in either case we can now weigh A plus B-upside-down or E plus F-upside-down as in the solution to the nominal-weight-known problem given above. Done, again, in four weighings.
if the odd coin is D or H, then ABCEFG will have weighed either $6x/4$ or $6y/4$ depending on which, so now we're actually done after only three weighings.
even if we don't know the nominal weight of the coins, we can identify the odd coin and which way up it goes with at most four weighings.
[What follows was written when I thought the question was about a two-arm balance scale of the traditional sort, rather than a weighing machine that weighs one set of things and tells you the total weight. If I've understood OP's clarifications right, this was not the intended question.]
any solution to the "standard" 12 coins puzzle (e.g., MA DO LIKE ME TO FIND FAKE COIN) will find the odd coin here in three weighings (just keep them all showing, say, heads throughout) and since that also tells you whether the different coin is heavy or light, it will tell you in this case whether the different coin is heavy-when-heads or light-when-heads.
And of course
you can't do it with fewer than three weighings because there aren't enough different possible outcomes from those weighings to distinguish the 24 different possibilities.