As it is written in the OP, the equality is not totally correct. Let me state and prove the following:
Proposition. Let $E \subset \mathbb R^n$ be a set of finite perimeter. For $\mathcal L^1$-a.e. $\rho>0$ the following equality holds: $$ P(E \cap B_{\rho}) = P(E, B_{\rho}) + \mathcal H^{n-1}(E \cap \partial B_{\rho}). $$
Proof.
Let $u \in BV(\mathbb R^n)$ and set $v_\rho:=u \chi_{B_{\rho}}$, being $\chi_A$ the characteristic function of a set $A$. Then we have
$$\tag{0}
Dv_{\rho} = (Du)\llcorner_{B_\rho} - \gamma_{\rho}(u) \cdot \nu \mathcal H^{n-1}\llcorner_{\partial B_{\rho}}
$$
where $\gamma_{\rho}(u)$ is the trace of $u$ on $B_\rho$. By a well-known characterization of the trace operator, we know that if $x \in \partial B_\rho$
$$\tag{1}
\lim_{r \to 0^+} \frac{1}{r^n} \int_{B_\rho \cap B_r(x)} \vert u(y)-\gamma_{\rho}(u)(x) \vert \, dy \to 0, \qquad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho.
$$
On the other hand, we know that $\mathcal L^n$-a.e. $x$ is a Lebesgue point for $u$, i.e.
$$\tag{2}
\lim_{r \to 0^+} \frac{1}{r^n} \int_{B_r(x)} \vert u(y)- u(x) \vert \, dy \to 0, \qquad \mathcal L^{n}\text{-a.e. } x \in \mathbb R^n.
$$
Furthermore, by Coarea Formula, we know that if $f$ is a Lipschitz map
$$
\vert \nabla f \vert \mathcal L^n = \mathcal H^{n-1}\llcorner_{\{f=t\}} \otimes \mathcal L^1.
$$
Now take $f=\vert \cdot \vert$ and hence
\begin{equation*}
\mathcal L^n = \mathcal H^{n-1}\llcorner_{\partial B_{\rho}} \otimes \mathcal L^1,
\end{equation*}
(which is the well-known formula for the polar change of coordinates). From this it follows that (2) is equivalent to
$$
\text{for } \mathcal L^1\text{-a.e. } \rho>0, \, \lim_{r \to 0^+} \frac{1}{r^n} \int_{B_r(x)} \vert u(y)- u(x) \vert \, dy \to 0, \quad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho.
$$
Hence, being the integrand non negative, we also deduce
$$\tag{3}
\text{for } \mathcal L^1\text{- a.e. } \rho>0, \lim_{r \to 0^+} \frac{1}{r^n} \int_{B_r(x) \cap B_\rho} \vert u(y)- u(x) \vert \, dy \to 0, \quad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho.
$$
By comparison between (1) and (3) we get
\begin{equation*}
\text{for } \mathcal L^1\text{ - a.e. } \rho>0, \quad \gamma_\rho(u) = u \vert_{\partial B \rho}, \quad \mathcal H^{n-1}\text{-a.e. } x \in \partial B_\rho.
\end{equation*}
Now it is easy to conclude: from (0) we have
$$
Dv_{\rho} = (Du)\llcorner_{B_\rho} - u \cdot \nu \mathcal H^{n-1}\llcorner_{\partial B_{\rho}}, \qquad \mathcal L^1 \text{-a.e. } \rho >0
$$
which is
$$
D(\chi_{E\cap B_{\rho}})= (D\chi_E)\llcorner_{B_\rho} - \chi_E \cdot \nu \mathcal H^{n-1}\llcorner_{\partial B_{\rho}}, \qquad \mathcal L^1 \text{-a.e. } \rho >0
$$
and since the two measures are mutually singular we deduce, taking total variations,
$$
\vert D(\chi_{E\cap B_{\rho}})\vert = \vert D\chi_E \vert \llcorner_{B_\rho} + \mathcal H^{n-1}\llcorner_{(\partial B_{\rho} \cap E)}, \qquad \mathcal L^1 \text{-a.e. } \rho >0.
$$