First, let's work out a standard problem, where $ g(w, t) = \mathbb{E} \left[ V(W_T) \middle \vert W_t = w \right] \, .$ Integrate both sides from $t$ to $T$: $ V(W_T) - g(W_t, t) = \int_t^T dg(W_s, s) = \int_t^T \left( \partial_s g \cdot ds + \partial_w g \cdot dW_s + \frac{1}{2} \partial_w^2 g \cdot ds \right) \, . $ Note that $\mathbb{E}\left[ \int_t^T \partial_w g \, dW_s \right] = 0$ since $dW_s$ is in the future of $W_t = w$. Next, take the expectation on both sides conditional on the filtration $\mathcal{F}_t$: $ \mathbb{E} \left[ V(W_T) - g(W_t, t) \middle \vert \mathcal{F}_t \right] = \mathbb{E} \left[ \int_t^T \left( \partial_s g + \frac{1}{2} \partial_w^2 g \right) ds \middle \vert \mathcal{F}_t \right] $
$ g(W_t, t) - g(W_t, t) = 0 = \mathbb{E} \left[ \int_t^T \left( \partial_s g + \frac{1}{2} \partial_w^2 g \right) ds \middle \vert \mathcal{F}_t \right] \, . $ We expect the integrand on the right-hand-side to be zero; that is, $ \partial_t g + \frac{1}{2} \partial_w^2 g = 0 \, , $ which is the backward PDE for $g(w,t)$.
Next, let's use the same approach for $f(x,\overline{x},t)$. Again, integrate both sides from $t$ to $T$: $ V(X_T) - f(X_t, \overline{X}_t, t) = \int_t^T df(X_s, \overline{X}_s, s) = \int_t^T \left( \partial_t f \cdot ds + \partial_x f \cdot dX_s + \frac{1}{2} \partial_x^2 f \cdot dX_s^2 + \partial_\overline{x} f \cdot d\overline{X}_s + \frac{1}{2} \partial_\overline{x}^2 f \cdot d\overline{X}_s^2 \right) \, . $ Note that $ dX_s^2 = \sigma^2 X_s^2 ds + \mathcal{O}(ds^2)$, and $d\overline{X}_s^2 = \lambda^2 \left(X_s - \overline{X}_s \right)^2 ds^2 = \mathcal{O}(ds^2) $.
Take the expectation on both sides conditional on the filtration $\mathcal{F}_t$: $ \mathbb{E} \left[ V(X_T) - f(X_t, \overline{X}_t, t) \middle \vert \mathcal{F}_t \right] = 0 = \mathbb{E} \left[ \int_t^T \left( \partial_t f + \partial_x f \cdot \mu X_s + \frac{1}{2} \partial_x^2 f \cdot \sigma^2 X_s^2 + \partial_\overline{x} f \cdot \lambda (X_s - \overline{X}_s) \right) ds \middle \vert \mathcal{F}_t \right] \, .$
Hence the backward PDE is $ \partial_t f + \partial_x f \cdot \mu X_s + \frac{1}{2} \partial_x^2 f \cdot \sigma^2 X_s^2 + \partial_\overline{x} f \cdot \lambda (X_s - \overline{X}_s) = 0 \, .$