= Solution
For $f(x)=x$, the <continuous martingale> is $M_t^f=B_t$. Since $B_t$ has the centered <normal distribution> of variance $t$,
$$
\mathbb E|M_t^f|^2=t=\int_0^t\mathbb E|f'(B_s)|^2\,ds.
$$
For $f(x)=x^2$, the <continuous martingale> is $B_t^2-t$. The <Gaussian fourth moment> gives
$$
\mathbb E(B_t^2-t)^2=3t^2-2t^2+t^2=2t^2=\int_0^t4\mathbb EB_s^2\,ds.
$$
This verifies both cases. The time integral in the printed identity is understood with its omitted $ds$.
Write $M^{(n)}=M^{f_n}$. The derivatives of the approximations are
$$
g_n(x):=f_n'(x)=\begin{cases}-1&x\le-1/n,\\nx&|x|<1/n,\\1&x\ge1/n.\end{cases}
$$
Thus $|g_n|\le1$. The <Itô formula> gives $M_t^{(n)}=\int_0^t g_n(B_s)\,dB_s$. This use of the formula is valid despite the two second-derivative discontinuities: mollify the $C^1$ function $f_n$, whose derivative is Lipschitz. Its smooth derivatives converge boundedly, while the second derivatives are bounded by $n$ and converge except at $\pm1/n$. For every $s>0$, the <normal distribution> of $B_s$ assigns those points zero <probability>. <Dominated convergence> for the drift integral and the <Itô isometry> for the stochastic integral pass the smooth formula to $f_n$. Equivalently, this identifies the <martingale> in the assumed decomposition and permits its cross-covariances to be calculated.
If $m\ge n$, then $|g_m-g_n|\le\mathbf1_{\{|x|<1/n\}}$. The <Itô isometry> and <Doob L2 maximal inequality> therefore give, for every fixed $T$,
$$
\mathbb E\sup_{s\le T}|M_s^{(m)}-M_s^{(n)}|^2\le4\int_0^T\mathbb E|g_m(B_s)-g_n(B_s)|^2\,ds\le4\int_0^T\mathbb P(|B_s|<1/n)\,ds.
$$
The <Gaussian density> is bounded by $(2\pi s)^{-1/2}$, so
$$
\mathbb P(|B_s|<1/n)\le\frac{\sqrt{2/\pi}}{n\sqrt s},\qquad \mathbb E\sup_{s\le T}|M_s^{(m)}-M_s^{(n)}|^2\le\frac{8\sqrt{2T/\pi}}n.
$$
Consequently the processes are Cauchy in the complete space of continuous random paths with norm $(\mathbb E\sup_{s\le T}|Z_s|^2)^{1/2}$. They have a continuous-path limit $M$ there. Limits obtained for different integer horizons agree on overlaps; choose these versions consistently. Each $M_t$ is adapted, and passing to the $L^2$ limit in $\mathbb E[M_t^{(n)}\mid\mathcal F_s]=M_s^{(n)}$ shows that $M$ is a <continuous martingale>. Letting $m\to\infty$ in the preceding bound proves the requested convergence, with an explicit rate:
$$
\boxed{\mathbb E\sup_{s\le T}|M_s^{(n)}-M_s|^2\le\frac{8\sqrt{2T/\pi}}n\longrightarrow0.}
$$
The same <Itô isometry> identifies $M_t=\int_0^t\operatorname{sgn}(B_s)\,dB_s$, with $\operatorname{sgn}(0)=0$.
There is a genuine \b[factor-of-two error in the last printed conclusion]. The stated $f_n$ has second derivative $n\mathbf1_{\{|x|<1/n\}}$ almost everywhere, and the generator in the decomposition is one half of the second derivative. Hence the exact identity is
$$
L_n(t):=\frac n2\int_0^t\mathbf1_{\{|B_s|<1/n\}}\,ds=f_n(B_t)-f_n(0)-M_t^{(n)}.
$$
Because $\sup_x|f_n(x)-|x||\le1/(2n)$ and $f_n(0)=1/(2n)$, this already proves uniform $L^2$ convergence on bounded intervals to $L_t=|B_t|-M_t$. It does not by itself prove almost-sure convergence of the full sequence, so we supply that additional step.
Take $n_k=k^2$. For each fixed integer $T$ and each $\varepsilon>0$, the maximal-error bound and <Markov's inequality> imply
$$
\sum_k\mathbb P\!\left(\sup_{s\le T}|M_s^{(k^2)}-M_s|>\varepsilon\right)\le\frac{8\sqrt{2T/\pi}}{\varepsilon^2}\sum_k\frac1{k^2}<\infty.
$$
The first <Borel-Cantelli lemma>, applied to rational positive $\varepsilon$ and integer $T$, gives uniform almost-sure convergence of this subsequence on every bounded interval. Therefore $L_{k^2}\to L$ uniformly there <almost surely>.
For $k^2\le n\le(k+1)^2$, the indicator windows are nested. The <ratio-one interpolation for shrinking-window occupation integrals> gives
$$
\frac n{(k+1)^2}L_{(k+1)^2}(t)\le L_n(t)\le\frac n{k^2}L_{k^2}(t).
$$
Both ratios tend uniformly to one as $k\to\infty$. The two bounds converge uniformly to the continuous nonnegative process $L$, so the entire sequence does too. Thus the correct normalized conclusion is
$$
\boxed{\frac n2\int_0^t\mathbf1_{\{|B_s|<1/n\}}\,ds\longrightarrow |B_t|-M_t\quad\text{almost surely},}
$$
and with the factor actually printed in the question the limit is
$$
\boxed{n\int_0^t\mathbf1_{\{|B_s|<1/n\}}\,ds\longrightarrow2(|B_t|-M_t).}
$$
These are the <occupation approximations for Brownian local time>. In the <Tanaka formula> convention, $L_t$ is the <local time of a semimartingale> at zero. It is not identically zero: $\mathbb EL_t=\mathbb E|B_t|=\sqrt{2t/\pi}>0$ for $t>0$, since $M$ starts at zero and is a <martingale>. Thus the discrepancy cannot be removed by treating the two limits as equal. The standard normalization is also given in https://galton.uchicago.edu/~lalley/Courses/385/ItoIntegral.pdf[Lalley's notes on the Itô calculus and local time].
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