Solve the chromatic puzzle
Octave, 334 313 bytes

Since the challenge may seem a bit daunting, I present my own solution. I did not formally prove that this method works (I guess that will come down to proving that the algorithm will never get stuck in a loop), but so far it works perfectly, doing 100x100 testcases within 15 seconds. Note that I chose to use a function with side effects rather than one that returns all the coordinates since that saved me a few bytes. Coordinates are row-major, 1-based, and formatted as row1 col1 row2 col2. Input colours are 0,1,2 since this works better with mod, at the cost of having to use numel rather than nnz. Golfed version: Edit: saved another few bytes by using a technique from Kevin Lau's answer.

function P(p)
k=0;[m,n]=size(p);t=@(v)mod(sum(v)*numel(v),3);for c=1:n
p(:,c)=V(p(:,c));end
k=1;for r=1:m
p(r,:)=V(p(r,:));end
function a=V(a)
while any(a-(g=t(a)))
isempty(q=find(diff(a)&amp;a(1:end-1)-g,1))&amp;&amp;(q=find(diff(a),1));
a([q q+1])=t(a([q q+1]));if k
disp([r q r q+1])
else
disp([q c q+1 c])
end;end;end;end


Example GIF of the solving algorithm:



Ungolfed version:

function solveChromaticPuzzle(p)
[m,n]=size(p);                           % Get size
t=@(v)mod(sum(v)*numel(v),3);            % Target colour function
for c=1:n                                % Loop over columns
    p(:,c)=solveVec(p(:,c));             % Solve vector
end
for r=1:m                                % Loop over rows
    p(r,:)=solveVec(p(r,:));
end
    function a=solveVec(a)               % Nested function to get globals
        g=t(a);                          % Determine target colour
        while(any(a~=g))                 % While any is diff from target...
            % Do the finding magic. Working left-to-right, we find the
            % first pair that can be flipped (nonzero diff) that does not
            % have the left colour different from our goal colour
            q=min(intersect(find(diff(a)),find(a~=g)));
            if(isempty(q))               % In case we get stuck...
                q=find(diff(a),1);       % ... just flip the first possible
            end;
            a([q q+1])=t(a([q q+1]));    % Do the actual flipping.
            if(exist('r'))               % If we're working per row
                disp([r q r q+1])        % Print the pair, using global row
            else
                disp([q c q+1 c])        % Print the pari, using global col
            end
        end
    end
end