To clarify your confusion:
Weierstrass M-test states one and only one thing:
Given a sequence of functions $f_k(x)$ defined on $E\subseteq\mathbb{R}$, the series $\sum f_k(x)$ converges uniformly if there exists a sequnece of reals $M_k$ such that $|f_k|\le M_k$ for each $k$ and $\sum M_k$ converges.
Note here that $M_k$ must not depend on $x$. This only means that if you found such sequence $M_k$ then the series uniformly converges. It says nothing about the series if you have found some $M_k$ whose series does not converge.
In particular, this test cannot be used to prove that some series does not converge uniformly.
Now consider
$$
f_k(x)=\frac{1}{kx+2}-\frac{1}{(k+1)x+2}
$$
Consider the partial sum
\begin{align*}
S_n(x)=\sum_{k=0}^{n} f_k(x)&=\left(\frac{1}{2}-\frac{1}{x+2}\right)+\cdots+\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right)\\
&=\frac{1}{2}-\frac{1}{(n+1)x+2}
\end{align*}
the series $\sum f_k(x)$ converges uniformly if and only if $S_n$ converges uniformly.
Now $S_n(0)=0$ for all $n$. So $S_n(0) \to 0$, and $S_n(x) \to \frac{1}{2}$ for $0<x\le 1$. So define
$$
S(x)=
\begin{cases}
0 & \text{if} \quad x=0 \\
\frac{1}{2} & \text{if} \quad 0<x\le 1
\end{cases}
$$
Then $S_n(x)\to S(x)$. Also, $S_n(0)-S(0)=0$ for any $n$. Now
$$
m_n:=\sup_{x\in [0,1]} |S_n-S|=\sup_{x\in (0,1]} \left\lvert -\frac{1}{(n+1)x+2} \right\rvert=\frac{1}{2} \not\to 0
$$
So $S_n$ does not converge uniformly.
Unlike $S_n(x) = \sum_{k=1}^n \sin kx$, the sum $\sum_{k=1}^n \frac{\sin kx}{k}$ is uniformly bounded for all $n$ and all $x \in \mathbb{R}$.
Hence, this series converges uniformly for all $x \in \mathbb{R}$ by the Dirichlet test -- since $(\ln k)^{-1} $ converges to $0$ monotonically and uniformly with respect to $x$.
Proving that $\sum_{k=1}^n \frac{\sin kx}{k}$ is uniformly bounded requires some effort. Because of periodicity, we can consider WLOG $x \in (0,\pi)$.
With $m = \lfloor1/x \rfloor$ we have
$$\left|\sum_{k=1}^n \frac{\sin kx}{k}\right| \leqslant \sum_{k=1}^{m}\frac{|\sin kx|}{k} + \left|\sum_{k=m+1}^{n}\frac{\sin kx}{k}\right| $$
For the first sum on the RHS,
$$\sum_{k=1}^{m}\frac{|\sin kx|}{k} \leqslant \sum_{k=1}^{m}\frac{k|x|}{k} = mx < 1$$
The second sum can be bounded as well using summation by parts.
Noting that $|S_n(x)| \leqslant \frac{1}{|\sin(x/2)|}$ and $|\sin(x/2)| \geqslant \frac{2}{\pi}\frac{x}{2} = \frac{x}{\pi} $ for $x \in (0,\pi)$ we have,
$$\left|\sum_{k=m+1}^{n}\frac{\sin kx}{k}\right| = \left|\frac{S_n(x)}{n} - \frac{S_m(x)}{m+1} + \sum_{k=m+1}^{n-1} S_k(x) \left(\frac{1}{k} - \frac{1}{k+1} \right)\right| \\ \leqslant \frac{2}{(m+1)|\sin(x/2)|} \\ \leqslant \frac{2\pi}{(m+1)x}\\\leqslant 2\pi$$
since $m = \lfloor 1/x \rfloor$ implies $(m+1)x \geqslant 1$.
Best Answer
If the series converges uniformly on $\mathbb{R}$, then for any $\epsilon > 0$ there exists $N_0 \in \mathbb{N}$ such that for all $N \geqslant N_0, x \in \mathbb{R}$ we would have
$$\left|x\sum_{n=N}^\infty \frac{\sin(n^2x)}{n^2}\right|= \left|x\sum_{j=0}^\infty \frac{\sin(N+j)^2x)}{(N+j)^2}\right|< \epsilon,$$
and it would follow that $\displaystyle \left|\sum_{j=0}^\infty \frac{\sin((N+j)^2x_q)}{(N+j)^2}\right| < \frac{\epsilon}{x_q}$ for all $q \in \mathbb{N}$ and $x_q = \frac{\pi}{2} + 2q\pi$.
Note that $\sin((N+j)^2x_q) = \sin\left((N+j)^2\frac{\pi}{2}\right)$ which equals $0$ if both $N$ and $j$ are odd and equals $\sin\left(N^2 \frac{\pi}{2}\right) = \pm 1$ if $N$ is odd and $j = 2k$ is even.
Hence, for any odd $N \geqslant N_0$ and all $q \in \mathbb{N}$ we have
$$\frac{\epsilon}{x_q} > \left|\sum_{j=0}^\infty \frac{\sin((N+j)^2x_q)}{(N+j)^2}\right| = \left|\sin\left(N^2 \frac{\pi}{2}\right)\right| \sum_{k=0}^\infty \frac{1}{(N+2k)^2} = \sum_{k=0}^\infty \frac{1}{(N+2k)^2} $$
This is impossible as the LHS converges to $0$ as $q \to \infty$ and the series on the RHS converges to a positive value. Therefore, the convergence is not uniform on $\mathbb{R}$.