**The answer is no.** Orthogonality is with respect to the $L^2$ norm and the $TV$ norm (of a smooth function) is the $L^1$ norm of the derivative and for this norm the $L^2$ orthogonality does not mean much.

Let $\Vert\cdot\Vert_1$ denote the $L^1$ norm on $\Omega$.
Smooth functions are dense in BV (Theorem 2, Section 5.2.2 in [1]) so I think you can approximate u by polynomials $Q_n$ and then you can write $P_n=Q_n-Q_{n-1}$. However, for smooth functions, and in particular for polynomials, $TV(P)=|DP|(\Omega)=\Vert \nabla P\Vert_1$.

Let $P$ and $Q$ be polynomials.
If $\nabla P$ is not parallel to $\nabla Q$ at every point of $\Omega$, then
$\Vert \nabla P + \nabla Q\Vert_1<\Vert \nabla P\Vert_1 + \Vert\nabla Q\Vert_1$.
Note that the gradients are parallel iff $\nabla P = \lambda\nabla Q$,
$\nabla (P-\lambda Q)=0$, $P=\lambda Q+c$ which is a very restrictive condition.
Therefore, unless you have this linear relation between all polynomials you have:
$$
TV(u)=|Du|(\Omega)=\lim_{k\to\infty}\Vert\sum_{n=1}^k \nabla P_n\Vert_1<\sum_{n=1}^\infty \Vert\nabla P_n\Vert_1=\sum_{n=1}^\infty TV(P_n).
$$
Therefore you cannot get the equality that you want, but instead you get a sharp inequality.

A good introduction to BV functions is Chapter 5 in:

[1] **L. Evans, R. F. Gariepy**, *Measure theory and fine properties of functions.* Revised edition. Textbooks in Mathematics. CRC Press, Boca Raton, FL, 2015.