= Solution
The intended setting is $X=M+A$ with $A$ continuous and adapted. These conditions are needed for the requested continuous-semimartingale conclusion; <finite variation> alone does not supply them. For example, $M=0$, $A_t=\mathbf1_{\{t\geq1\}}$ satisfies the printed deterministic bound and starts at zero, but $f_n(X)$ has a jump for sufficiently large $n$. We use the continuous adapted interpretation throughout this question.
Define $f_n(x)=\int_0^x\phi(ny)dy$. Then $f_n\in C^2$, $|f_n'|\leq1$ and $f_n''(x)=n\phi'(nx)\geq0$. The <Itô formula> gives the canonical <continuous semimartingale decomposition>
$$
\boxed{f_n(X_t)=N_t^{(n)}+V_t^{(n)},\quad N_t^{(n)}=\int_0^t f_n'(X_s)dM_s,\quad V_t^{(n)}=\int_0^t f_n'(X_s)dA_s+\frac12\int_0^t f_n''(X_s)d[M]_s.}
$$
The initial value is zero. The first term is a <continuous martingale>, since $M$ is bounded and the integrand is bounded. The second is continuous adapted <finite variation>, hence predictable. This decomposition is unique because a <continuous finite-variation local martingale is constant>. The phrase Doob-Meyer here must mean this <semimartingale> decomposition: $V^{(n)}$ need not be nondecreasing, and the hypotheses do not make $f_n(X)$ a submartingale as required by the <Doob-Meyer decomposition theorem>. Indeed $f_n'(0)=-1$, so the bounded deterministic drift $A_t=t\wedge\delta$ makes $f_n(A_t)$ initially decrease, excluding the submartingale property.
The sign process is predictable because it is a Borel function of the continuous adapted, and therefore predictable, process $X$. Its paths need not themselves be left-continuous. Pointwise, including at $x=0$,
$$
f_n'(x)=\phi(nx)\longrightarrow\operatorname{sgn}_-(x):=\mathbf1_{\{x>0\}}-\mathbf1_{\{x\leq0\}},\qquad|f_n'(x)-\operatorname{sgn}_-(x)|\leq2.
$$
For a fixed horizon $T$, $\mathbb E[M]_T=\mathbb EM_T^2\leq K^2$. The <dominated convergence theorem> for the finite measure $\mathbb E\,d[M]$ on $[0,T]$ gives
$$
\mathbb E\int_0^T|f_n'(X_s)-\operatorname{sgn}_-(X_s)|^2d[M]_s\longrightarrow0.
$$
The <Itô isometry>, extended from simple predictable integrands by their density, and the <Doob L2 maximal inequality> now imply
$$
\mathbb E\sup_{t\leq T}\left|N_t^{(n)}-\int_0^t\operatorname{sgn}_-(X_s)dM_s\right|^2\leq4\mathbb E\int_0^T|f_n'(X_s)-\operatorname{sgn}_-(X_s)|^2d[M]_s\longrightarrow0.
$$
Hence
$$
\boxed{N^{(n)}\longrightarrow\operatorname{sgn}_-(X)\cdot M\text{ uniformly on compacts in probability}.}
$$
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