I suggest a deterministic solution using some basic TM trickery.
First you need to identify the parts. Suppose we tag the 0,1 symbols by replacing them by symbols, e.g. A,B (A~0, B~1) - so that we know what we already "tagged" but keep the data "intact". The TM works in a simple loop: move far left, tag the first untagged element, move to far right, tag first two untagged elements. If at some point the tags from left and the tags from right meet, we have found the end of W prefix. If the tags don't meet well, then the length of the input is not divisible by 3 and the TS says NO. Suppose we used A,B tags on left and C,D tags on right (so we can always find the end of W prefix). We split in the C,D part into WR and W in a similar way.
Suppose that the first part now consist of A,B symbols, the middle of C,D symbols and the last of E,F symbols. The TS now reads the three sections in parallel (while the middle one is read in reverse) and compares them in a simple loop, while rewriting the correct parts with some special # symbol. First, move to the left to find the first non-# symbol (which is either A or B). Remember it (use two sets of states, one corresponding to A and one corresponding to B) and replace the symbol by #. Move right until you reach the end of C,D section (the first symbol E,F or # signifies the end), and check if the previous symbol corresponds to the remembered value (A~C, B~D). If it does not, answer NO. Otherwise replace it by # and move to the first E/F-symbol and again check (A~E,B~F) and replace with #. Repeat until there are no non-# symbols left. Once that happens, the answer is YES.