The sum is clearly bounded for $a = 1, f(k) = 0$, but what about others?
Using Mathematica software, I observed that the partial sums $\sum_{k=1}^{n} \sin(k + \log k)$ and $\sum_{k=1}^{n} \sin(k + \sin k)$ might be bounded, as they appeared so for $n$ up to $10^7$ (however, this doesn't actually prove anything).
What are the conditions of $a \in \mathbb{R}$ and $f(k)= o(k)$ that would make $\sum_{k=1}^{n} \sin(a k + f(k))$ bounded?
When $a \in \pi \mathbb{Q}$, I could rewrite $a$ as $a = \frac{p}{q} \pi$, where $p$ and $q$ integers that are coprime. Then:
\begin{aligned} \sum_{k=1}^{n} \sin(a k + f(k)) &= \sum_{k=1}^{n} \sin(\frac{p}{q} \pi k + f(k)) \\ &= \sum_{a=1}^{2q} \sum_{b=0}^{\lfloor \frac{n - a}{2q} \rfloor} \sin(\frac{(a + 2 q b) p \pi}{q} + f(a + 2 q b)) \\ &= \sum_{a=1}^{2q} \sum_{b=0}^{\lfloor \frac{n - a}{2q} \rfloor} \sin(\frac{a p \pi}{q} + f(a + 2 q b)) \end{aligned}
Then the above is a sum of some $\sum \sin(g(k))$ with $g(k) = o(k)$, so I can consider them individually.
What if $a \not \in \pi \mathbb{Q}$? That is, when $a k$ is equidistributed modulo $2 \pi$?
Not a very good bound
$$\sum \sin(ak+f(k))=\sum \sin(ak)\cos(f(k))+\sum \cos(ak)\sin(f(k))$$
If $f(k)=C+o(1)$, then $$\cos f(k)=\cos C\cos o(1)-\sin C\sin o(1)=\cos C+o(1)$$
$$\sin f(k)=\sin C\cos o(1)+\cos C\sin o(1)=\sin C+o(1)$$
By knowing that $\sum^n\sin ak$ and $\sum^n\cos ak$ is bounded, the original sum is bounded as well.