Though the solution of both are working, there are also other methods. The method of Yehuda is recursive which is not that nice, and can therefore not be parallelised easily. I see wo other ways of doing it:

ClearAll[distance];
distance[{{x1_,y1_},{x2_,y2_}}]:=Abs[(x2-x1)]+Abs[(y2-y1)];
totaldistance[x:{{_,_}..}]:=Total[distance/@Partition[x,2,1,{-1,1},{{0,0}}]]

similar to the above 2 replies. But one can also do it like this:

ClearAll[distance];
ClearAll[distance];
distance[{\[CapitalDelta]x_,\[CapitalDelta]y_}]:=Abs[\[CapitalDelta]x]+Abs[\[CapitalDelta]y];
totaldistance[x:{{_,_}..}]:=Total[distance/@Differences[Join[{{0,0}},x,{{0,0}}]]]

or even:

ClearAll[totaldistance];
totaldistance[x:{{_,_}..}]:=Total[Abs[Differences[Join[{{0.0,0.0}},x,{{0.0,0.0}}]]],\[Infinity]]

Note that prepending/appending 0.0 instead of 0 can make a big difference depending on the input data. Especially if it is a packed array of a certain type. The last one will be the fastest as it does not use Map, and everything is vectorized (assuming real numbers, not symbolic calculation, and 'large' number of points).

You can use Partition to divide your path into couples of successive points, and then map distance on the couples:

Clear[distance];
distance[{{x1_, y1_}, {x2_, y2_}}] := Abs[x2 - x1] + Abs[y2 - y1];
distance[{pts__List}] := 
 Total[Map[distance, Partition[{pts}, 2, 1, 1]]]

The parameters 2, 1, 1 given to Partition mean that the sublists will be of length 2, with overhang 1, and the last point will appear as the first element in the last couple, with the second element taken from the beginning of the path. It's complicated. I hope I didn't mix it up.