Here's how I'd do it. Starting with diffcoord

diffcoord = {{2.87796, 29.2153}, {-13.6571, -19.1573}, {2.9201, 
  8.7201}, {3.52192, -16.1424}, {0.53561, 19.1832}, {3.76663, 
  3.32865}, {-11.0295, -10.1585}, {14.9637, 
  19.0262}, {-12.4468, -28.5079}};

Pair each element with the next and Dot the pairs.

Dot@@@(Transpose@{diffcoord[[1;;-2]],diffcoord[[2;;-1]]})

The output is

{-598.991, -206.934, -130.479, -307.777, 65.8716, -75.3581, -358.32,
-728.647}

There is certainly a more clever way to combine the two if you need to do this on huge lists, but this makes sense to me.

Here's how I'd do it. Starting with diffcoord

diffcoord = {{2.87796, 29.2153}, {-13.6571, -19.1573}, {2.9201, 
  8.7201}, {3.52192, -16.1424}, {0.53561, 19.1832}, {3.76663, 
  3.32865}, {-11.0295, -10.1585}, {14.9637, 
  19.0262}, {-12.4468, -28.5079}};

Pair each element with the next and Dot the pairs.

Dot@@@(Transpose@{diffcoord[[1;;-2]],diffcoord[[2;;-1]]})

The output is

{-598.991, -206.934, -130.479, -307.777, 65.8716, -75.3581, -358.32,
-728.647}

There is certainly a more clever way to combine the two if you need to do this on huge lists, but this makes sense to me.

Edit: @Guess who it is reminded me that Partition has a 3rd argument to specify an offset, which makes this a lot nicer looking!

Dot@@@Partition[diffcoord,2,1]