Addendum
There is a potential flaw in the original proof (below) which assumes that convergence to right-hand limits is uniform. Here is a different proof.
Any function of bounded variation has at most countably many jump discontinuities. There are two possibilities.
(1) There are finitely many jump discontinuities in some open neighborhood of $\lambda$. In this case, there is an interval $(\lambda - \delta, \lambda + \delta)$ where $f$ is continuous except possibly at $\lambda$. For $x \neq \lambda$ we have $f_R(x) = f(x+) = f_L(x) = f(x-) = f(x).$ By hypothesis $f_R$ is continuous at $\lambda$ and, therefore, $f_R(\lambda-) = f_R(\lambda+)$.
Thus $$f(\lambda-) = \lim_{x \to \lambda -}f(x) = \lim_{x \to \lambda -}f_R(x) = f_R(\lambda-) = f_R(\lambda+) = \lim_{x \to \lambda +}f_R(x) = \lim_{x \to \lambda +}f(x) = f(\lambda+),$$
and $f$ is continuous at $\lambda.$
(2) Jump discontinuities accumulate at $\lambda$. Let $(y_n)$ and $(z_n)$ be any increasing and decreasing sequences of points, respectively, both converging to $\lambda$. Since, the sums of the jumps must be bounded by the total variation of $f$, we have
$$\sum_{k=1}^\infty |f_R(y_k) - f_L(y_k)| < \infty, \\ \sum_{k=1}^\infty |f_R(z_k) - f_L(z_k)| < \infty,$$
and $\lim_{k \to \infty} |f_R(y_k) - f_L(y_k)| = \lim_{k \to \infty} |f_R(z_k) - f_L(z_k)| = 0$.
Thus,
$$f_L(\lambda-) = \lim_{k \to \infty} f_L(y_k) = \lim_{k \to \infty} f_R(y_k) - \lim_{k \to \infty} (f_R(y_k) - f_L(y_k)) = f_R(\lambda-), \\ f_L(\lambda+) = \lim_{k \to \infty} f_L(z_k) = \lim_{k \to \infty} f_R(z_k) - \lim_{k \to \infty} (f_R(z_k) - f_L(z_k)) = f_R(\lambda+), $$
and
$$f_L(\lambda-) = f_L(\lambda+) = f_R(\lambda+) = f_R(\lambda-).$$
Original Proof
By hypothesis, $\tilde{f}$ is continuous at $\lambda$. We must have $f(\lambda-) = f(\lambda+) =\tilde{f}(\lambda)$ when $ f$ is continuous at $\lambda$.
Suppose $f$ is not continuous at $\lambda$. This implies $f(\lambda-) \neq f(\lambda+) = \tilde{f} (\lambda).$ Since the left-hand limit exists there exists a sequence $(x_n)$ converging to $\lambda$ from the left such that $f(x_n)$ does not converge to $\tilde{f}(\lambda)$.
Hence, there exists $\epsilon_0 > 0$ such that for all sufficiently large $n$,
$$|f(x_n) - \tilde{f}(\lambda)| > \epsilon_0.$$
Now choose a sequence of points $(y_n)$ such that $x_n - 1/2^{n} < y_n < x_n.$ Then $y_n \to \lambda$ and $|f(x_n) - \tilde{f}(y_n)| < \epsilon_0/2$ for sufficiently large $n$, since the right-hand limit exists everywhere (and convergence is uniform in a compact neighborhood of $\lambda$).
Thus,
$$|\tilde{f}(y_n) - \tilde{f}(\lambda)| > |f(x_n) - \tilde{f}(\lambda)| - |f(x_n) - \tilde{f}(y_n)| > \epsilon_0/2, $$
which contradicts the continuity of $\tilde{f}$ at $\lambda.$
Best Answer
The main crude intuition behind BV one dimensional maps is that on a compact interval you want to have “finite variation”, that roughly means the map “cannot oscillate too much”: you cannot be able to find a sequence of finite partitions on the x-axis such that the image values of the sequence of partitions go to infinity.
Note that a function of bounded variation (BV henceforth) in one dimension may not be continuous and a continuous function may not be BV on a compact interval. For the counterexamples just take any monotone function on a closed interval and the following function defined on $[0,1]$ $$ u(x)=x^a\sin(\frac{1}{x^b}) $$ defined to be zero at zero is not of bounded variation as long as $1 \le a\leq b$.
Nevertheless, BV functions have nice differentiability properties, at least if you are willing to accept throwing away a “measure theoretic” small set of points. Indeed, a function of bounded variation on a compact interval is necessary differentiable a.e., so that any continuous function which is not differentiable on a positive measure set is not of bounded variation.
Also, some function which oscillates too much can not be bounded variation as the example above shows.
So, the notions of continuity and BV are kind of transversal: the best way to think about it is in terms of “controlled oscillations” on a compact set.