Coefficient list of an expression with indices

Yet another way to do this, using Coefficient

expr = -27 Subscript[u, 1, 1] - 6 Subscript[u, 1, 2] + 
   9 Subscript[u, 1, 3] + 36 Subscript[u, 2, 1] + 
   8 Subscript[u, 2, 2] - 12 Subscript[u, 2, 3] - 
   9 Subscript[u, 3, 1] - 2 Subscript[u, 3, 2] + 
   3 Subscript[u, 3, 3];
Flatten@Array[Coefficient[expr, Subscript[u, #1, #2]] &, {3, 3}]
(* {-27, -6, 9, 36, 8, -12, -9, -2, 3} *)

Edit

Here's a way to get there using CoefficientList,

CoefficientList[expr, Variables[expr]] // Flatten // 
   DeleteDuplicates // Reverse // Most
(* {-27, -6, 9, 36, 8, -12, -9, -2, 3} *)

Since all terms have Subscript[something], we can use a pattern:

expr /. Plus -> List /. c_.*Subscript[__] :> c

{-27, -6, 9, 36, 8, -12, -9, -2, 3}

This assumes the expression has been Expanded so that the Head is indeed Plus. This will also give the coefficients in the same order as they appeared in the expression. If you also would like which coefficients belong with which coefficients, do:

expr /. Plus -> List /. c_.*Subscript[x__] :> {c, Subscript[x]}

{{-27, Subscript[u, 1, 1]}, {-6, Subscript[u, 1, 2]}, {9, Subscript[u, 1, 3]}, {36, Subscript[u, 2, 1]}, {8, Subscript[u, 2, 2]}, {-12, Subscript[u, 2, 3]}, {-9, Subscript[u, 3, 1]}, {-2, Subscript[u, 3, 2]}, {3, Subscript[u, 3, 3]}}

EDIT:

Explanation of replacement: First, all terms in expr are put in a list. Then, we replace all instances of expressions in that list that match the pattern c_.*Subscript[__] with just c. This pattern matches expressions where some expression c is multiplied by a Subscript expression with any arguments (at least one). The period in the pattern c_. lets us include the default value in multiplication, which is 1. If we had omitted it, the pattern would not match e.g. the last term in

3 Subscript[u,1,1] + Subscript[u,1,2]

because that term is not strictly "some expression multiplied with Subscript.

As for the ordering, yes, the ordering will be the same as that in expr as long as it is a "flat" sum, i.e. fully expanded.

Turning my comment into an answer. If you know that this is exactly the form of the expression, you can just pick out the coefficients manually with First. It's important though that you first turn the sum into a list, because otherwise the coefficients will be summed up immediately:

First /@ List @@ expression

The List @@ replaces the head Plus with List. Then we simply pick out each coefficient with First, because Mathematica orders products with the coefficient first and the variable second.

Note that this doesn't work reliably if some of the terms have a coefficient of 1 (in that case, the variable itself would be returned). You could fix that with something like:

Replace[First /@ List @@ expression, n_ /; ! NumericQ@n -> 1, {1}]

But at that point it might be simpler to use one of the other answers.

Alternatively, as garej suggested in a comment, if you have a variable a that you know is unused, you could do:

First /@ List @@ Expand[a*expr] /. a -> 1