Something that might prove applicable is the Dirichlet Convergence Test. Summation by Parts yields
$$
\sum_{k=1}^na_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\tag{1}
$$
where $\displaystyle A_n=\sum_{k=1}^na_k$. Note that $(-1)^{n+1}\sin(nx)=-\sin(n(\pi+x))$, so we have
$$
\begin{align}
A_n
&=\sum_{k=1}^n(-1)^{n+1}\sin(kx)\\
&=-\sum_{k=1}^n\sin(k(\pi+x))\\
&=\frac{\sin(n(\pi+x)/2)}{\sin((\pi+x)/2)}\sin((n+1)(\pi+x)/2)\tag{2}
\end{align}
$$
Thus, $|A_n|\le|\sec(x/2)|$. Therefore,
(i) according to $(1)$ and $(2)$, the series converges pointwise except at odd multiples of $\pi$; however, at any multiple of $\pi$, each term is $0$. Thus, the series converges for all $x$.
(ii) according to $(1)$ and $(2)$, the series converges uniformly on compact sets not containing an odd multiple of $\pi$. A sequence of continuous functions cannot converge uniformly to a discontinuous function, so the convergence cannot be uniform in any neighborhood of an odd multiple of $\pi$.
(iii) except at multiples of $\pi$, the series is not absolutely convergent. Here is the outline of a proof.
First, show that if $x$ is not a multiple of $\pi$, at least $1/4$ of the multiples of $x$ have $|\sin(kx)|>1/\sqrt{2}$. Because $|\sin(k(x+n\pi))|=|\sin(kx)|$, we only need to consider $x\in[-\pi/2,\pi/2]$. Since $|\sin(-kx)|=|\sin(kx)|$, we only need to consider $x\in[0,\pi/2]$.
If $x\in[\pi/(n+1),\pi/n]$, then at least $\lfloor n/2\rfloor$ multiples of $x$ are in $[(m+1/4)\pi,(m+3/4)\pi]$ out of at most $n+1$ in $[m\pi,(m+1)\pi]$. Since $(0,\pi/2]=\cup_{n=2}^\infty[\pi/(n+1),\pi/n]$, this means that at least $1/4$ of the multiples of $x$ in $[m\pi,(m+1)\pi]$ are in $[(m+1/4)\pi,(m+3/4)\pi]$, and therefore, have $|\sin(kx)|>1/\sqrt{2}$.
Thus, $1/4$ of each $n$ to $n+1$ consecutive terms of the sum have $|\sin(kx)|>1/\sqrt{2}$. The sum is therefore greater than
$$
\frac{1}{4}\sum_{k=n+1}^\infty\frac{2}{k}\frac{1}{\sqrt{2}}\tag{3}
$$
which diverges. Thus, if $x$ is not a multiple of $\pi$, the series of absolute values diverges.
The function
$$
g(x)= \int_0^x f(t)dt-\frac{x}{2\pi}\int_0^{2\pi}f(t)dt
$$
is periodic and continuous on $[0,2\pi]$, and is of bounded variation on $[0,2\pi]$ because $f$ is bounded (assuming $f$ is Riemann integrable, as you indicated in your comments.) Therefore, the Fourier series for $g$ converges uniformly to $g$ on $[0,2\pi]$.
Best Answer
It can be proved, that Fourier series $S(x)$ of piecewise smooth function on $(-\pi, \pi)$ converges pointwise to the following value:
$$ \begin{aligned} S(x) &= \frac{f(x+0)+f(x-0)}{2},\ \forall x \in (-\pi, \pi)\\ S(-\pi) = S(\pi) &= \frac{f(\pi-0)+f(-\pi+0)}{2} \end{aligned} $$
Fortunately, $f(x+0) = f(x-0) = f(x),\ \forall x \in (-\pi, \pi)$, so that
$$ S(x) = f(x), \forall x \in (-\pi, \pi) $$
Since $f(\pi-0) = f(-\pi+0) = 0$
$$ \begin{aligned} S(\pi) &= 0 = f(\pi)\\ S(-\pi) &= 0 = f(-\pi) \end{aligned} $$
Because of periodicity
$$ S(x) = f(x),\ \forall x \in R $$