Let $H$ be a $\mathbb R$-Hilbert space and $H_\lambda$ be a closed subspace of $H$ for $\lambda\in\mathbb R$. Assume $H_\lambda\subseteq H_\mu$ for all $\lambda,\mu\in\mathbb R$ with $\lambda\le\mu$ and let $\pi_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$ for $\lambda\in\mathbb R$. Moreover, let $$\varrho_x(\lambda):=\left\|\pi_\lambda x\right\|_H^2\;\;\;\text{for }\lambda\in\mathbb R$$ and $$\operatorname P_x(\lambda):=\pi_\lambda x\;\;\;\text{for }\lambda\in\mathbb R$$ for $x\in H$.
Fix $x\in H$. Are we able to show that the variation of $\varrho_x$ and $\operatorname P_x$ on any interval coincides?
In order to prove the desired claim, let $k\in\mathbb N$ and $\lambda_0,\ldots,\lambda_k\in\mathbb R$ with $\lambda_0\le\cdots\le\lambda_k$. We need to show that $$A:=\sum_{i=1}^k\left|\varrho_x(\lambda_i)-\varrho_x(\lambda_{i-1})\right|=\sum_{i=1}^k\left\|\operatorname P_x(\lambda_i)-\operatorname P_x(\lambda_{i-1})\right\|_H=:B.$$ Since $\varrho_x$ is nondecreasing, $A=\varrho_x(\lambda_k)-\varrho_x(\lambda_0)$. Moreover, we easily see that $$\left\|\operatorname P_x(\lambda_i)-\operatorname P_x(\lambda_{i-1})\right\|_H^2=\varrho_x(\lambda_i)-\varrho_x(\lambda_{i-1})\tag1$$ for all $i\in\left\{1,\ldots,k\right\}$.
So, it seems like the claim is wrong, if I'm not missing something. I would at least like show that the variation of $\operatorname P_x$ is bounded by the variation of $\varrho_x$, i.e. $B\le A$, but that seems to be wrong too.
Remark: The question came up to my mind as I was considering the construction of the spectral measure for self-adjoint operators on a Hilbert space. In the construction of that measure, one is basically integrating against a right-continuous function of bounded variation of the form $\operatorname P_x$, but one is usually defining the domain of the integration to be $L^2(\varrho_x)$. So, it seems like one is thinking that $\varrho_x$ and $\operatorname P_x$ (which gives rise to a Lebesgue-Stieltjes vector measure) have the same variation.
Let's build a simple example. Let $H = \ell^2(\mathbb{N})$ and let $H_\lambda = \operatorname{span}\{ e_i: 1 \leq i \leq \lambda\}$ where $\{e_i: i \in \mathbb{N}\}$ is the usual orthonormal basis of $\ell^2(\mathbb{N})$. Then we have that $(\pi_\lambda x)_k = x_k 1_{\{k \leq \lambda\}}$.
Now for simplicity suppose we are interested in the variation over the interval $[\frac12, \frac32]$. This isn't essential but will make the example simpler to see here.
Then, as you note, we have that $A = \rho_x(\frac32) - \rho_x(\frac12)$ so that $A = |x_1|^2$. Now, $B = \|P_x(\lambda_i) - P_x(\lambda_{i-1})\|$ for the unique value of $i$ such that $\lambda_{i-1} < 1 \leq \lambda_i$. Hence $B = |x_1|$.
This means you can't expect to have $A \lesssim B$ or $B \lesssim A$ in general. Indeed, if we were to assume either inequality was true, we could get a contradiction by considering the above example applied to one of the sequences of elements of $\ell^2$ given by $x^{(n)} = ne_1$ and $y^{(n)} = \frac1n e_1$.