consider a differentiable function $f: \mathbb{R}^n \rightarrow \mathbb{R} $ that attains a global minimum at $x$. Consider the following limiting direction with $t_k \to 0$ and $z_k \to x$ for $k \to \infty$
$$d = \lim_{{k \to \infty}} \frac{{z_k - x}}{{t_k}}$$
This means:
$$ z_k = x + t_k d + o\left(t_k \right) $$
Since in $x$ is a global minimum, we have:
$$ 0 \leq \frac{1}{t_k} (f(z_k ) - f(x)) = \frac{1}{t_k} \nabla f(x)^\top t_kd + \frac{1}{t_k} o(t_k)$$
My question is why the last term with the o notation $ \frac{1}{t_k} o(t_k)$ does not change? Is the last equality some kind of Taylor expansion and the remainder is somehow merged into the o-term?
Yes. Take into account that $o(\cdot)$ is a qualitative notation, so $o(\cdot)+o(\cdot)=o(\cdot)$ and $o(k(\cdot))=o(\cdot)$ etc.
By Taylor expansion with remainder in the form of Peano, $$ f(z_k)-f(x)=\nabla f(x)^T(z_k-x)+o(z_k-x)=\nabla f(x)^T(z_k-x)+o(t_k) $$ But you also have $$ \nabla f(x)^T(z_k-x)=\nabla f(x)^Tt_kd+\nabla f(x)^To(t_k)=\nabla f(x)^Tt_kd+o(t_k). $$ Putting everything together and dividing throughout by $t_k$ you get what you need.