I want to see that $B=k[t]_{(t)}$ is a valuation ring. To show this I need to show that for every $x\in \operatorname{Frac}(B)$ we have at least one of $x\in B$ or $x^{-1}\in B$. Intuitively i think $\operatorname{Frac}(B)$ should be $k(t)$ but I can't think of succinct way of showing this.
Let $x\in k(t)$ then $x=f/g$ with $f,g\in k[t]$. An element $y\in B$ looks like $h/t^n$. So we need to show that $f/g=h/t^n$ or $g/f=h/t^n$. I think we I should try use that $k[t]$ is a UFD and get a nice relation between $f$ and $g$. We can assume $f,g$ to be coprime,but even them I'm not sure ho we can deal with something like this $x=\frac{t-1}{t-2}$. If we could show that that $t$ must divide $f$ or $g$ we would be done but I am not sure if this is true.
Questions: Why is $\operatorname{Frac}(B)=k(t)$? Why can we assume $t$ must divide $f$ or $g$?
$k[t]_{(t)}$ is a localization of $k[t]$, so it is canonically a subring of $\operatorname{Frac} k[t]=k(t)$. If $F$ were any field containing $k[t]_{(t)}$ properly contained in $k(t)$, this would contradict the fact that $k[t]$ has fraction field $k(t)$.
Your second problem has a key misunderstanding: $k[t]_{(t)}$ is defined to be the localization at the set of elements not in $(t)$, not the localization at $\{1,t,t^2,t^3,\cdots\}$. So $x$ is already in $k[t]_{(t)}$, for instance. Try again with this new information - I've put a full explanation under a spoiler below for you.