Solution:

Let $O_{i}$ be the centre of $\omega_{i}$, $i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo $4$), to infer that $\omega_{4}$ is internally tangent to $\omega_{1}$ at $T_{4}$, and $O_{1} O_{2} O_{3} O_{4}$ is a (possibly degenerate) parallelogram.

Let $F_{i}$ be the foot of the perpendicular from $O_{i}$ to $PW$. The $F_{i}$ clearly bisect the segments $PW$, $QR$, $ST$ and $UV$, respectively.

The proof can now be concluded in two similar ways.

First Approach. Since $O_{1} O_{2} O_{3} O_{4}$ is a parallelogram, $\overrightarrow{F_{1} F_{2}} + \overrightarrow{F_{3} F_{4}} = \mathbf{0}$ and $\overrightarrow{F_{2} F_{3}} + \overrightarrow{F_{4} F_{1}} = \mathbf{0}$; this still holds in the degenerate case, for if the $O_{i}$ are collinear, then they all lie on the line $T_{1} T_{4}$, and each $O_{i}$ is the midpoint of the segment $T_{i} T_{i+1}$. Consequently,

$$\begin{aligned} \overrightarrow{PQ} - \overrightarrow{RS} + \overrightarrow{TU} - \overrightarrow{VW} = & \left(\overrightarrow{P F_{1}} + \overrightarrow{F_{1} F_{2}} + \overrightarrow{F_{2} Q}\right) - \left(\overrightarrow{R F_{2}} + \overrightarrow{F_{2} F_{3}} + \overrightarrow{F_{3} S}\right) \\ & + \left(\overrightarrow{T F_{3}} + \overrightarrow{F_{3} F_{4}} + \overrightarrow{F_{4} U}\right) - \left(\overrightarrow{V F_{4}} + \overrightarrow{F_{4} F_{1}} + \overrightarrow{F_{1} W}\right) \\ = & \left(\overrightarrow{P F_{1}} - \overrightarrow{F_{1} W}\right) - \left(\overrightarrow{R F_{2}} - \overrightarrow{F_{2} Q}\right) + \left(\overrightarrow{T F_{3}} - \overrightarrow{F_{3} S}\right) - \left(\overrightarrow{V F_{4}} - \overrightarrow{F_{4} U}\right) \\ & + \left(\overrightarrow{F_{1} F_{2}} + \overrightarrow{F_{3} F_{4}}\right) - \left(\overrightarrow{F_{2} F_{3}} + \overrightarrow{F_{4} F_{1}}\right) = \mathbf{0} \end{aligned}$$

Alternatively, but equivalently, $\overrightarrow{PQ} + \overrightarrow{TU} = \overrightarrow{RS} + \overrightarrow{VW}$, as required.

Second Approach. This is merely another way of reading the previous argument. Fix an orientation of the line $PW$, say, from $P$ towards $W$, and use a lower case letter to denote the coordinate of a point labelled by the corresponding upper case letter.

Since the diagonals of a parallelogram bisect one another, $f_{1} + f_{3} = f_{2} + f_{4}$, the common value being twice the coordinate of the projection to $PW$ of the point where $O_{1} O_{3}$ and $O_{2} O_{4}$ cross; the relation clearly holds in the degenerate case as well.

Plug $f_{1} = \frac{1}{2}(p + w)$, $f_{2} = \frac{1}{2}(q + r)$, $f_{3} = \frac{1}{2}(s + t)$ and $f_{4} = \frac{1}{2}(u + v)$ into the above equality to get $p + w + s + t = q + r + u + v$. Alternatively, but equivalently, $(q - p) + (u - t) = (s - r) + (w - v)$, that is, $PQ + TU = RS + VW$, as required.