It’s a straightforward result that a uniform space $X$ is totally bounded iff every net in $X$ has a Cauchy subnet, and the usual equivalence between nets and filters shows that this is in turn equivalent to the statement that for each filter in $X$ there is a finer Cauchy filter. However, this does not guarantee that every sequence has a Cauchy subsequence. One counterexample is $\beta\omega$: it’s a compact Hausdorff space, so it has a unique compatible uniformity and is both complete and totally bounded in that uniformity, but the sequence $\langle n:n\in\omega\rangle$ has no convergent subsequence.
I’ve not worked much with uniform spaces, but in that context I’ve generally seen precompact used as a synonym of totally bounded. The situation with respect to the other meaning of the term is the same as in metric spaces. If $Y$ is a subspace of a complete uniform space $X$, then $\operatorname{cl}_XY$ is complete; if $Y$ is also totally bounded $-$ which is an inherent property, just as it is for metric spaces $-$ then $\operatorname{cl}_XY$ is totally bounded and therefore compact. Thus, a subset of a complete uniform space is precompact (in this other sense) iff it is totally bounded. This will not be the case in arbitrary uniform spaces, however.
I already showed in this answer that $X'$ relatively compact implies $X'$ relatively countably compact. Also, if $X'$ is relatively compact, the $\overline{X'}$ is compact, and thus sequentially compact (this holds in particular in metric spaces, but also more broadly). So any sequence from $X'$ has a convergent subsequence with limit in $\overline{X'}$, so $X'$ is then relatively sequentially compact as well.
In any space, $X'$ relatively sequentially compact implies $X'$ relatively countably compact: any infinite subset $A$ of $X'$ contains some sequence with all different elements, which has a convergent subsequence to some $x \in X$, and this $x$ is an accumulation point of $A$.
If $X'$ is relatively countably compact, and $X$ is metric (first countable and $T_1$ will already do), let $(x_n)$ be a sequence from $X'$. If $A = \{x_n: n \in \mathbb{N}\}$ is finite, some value occurs an infinite number of times, and yields a convergent subsequence. So assume $A$ is infinite, so it has an accumulation point $p \in X$. Because $X$ is $T_1$, this means that every neighbourhood of $p$ intersects $A$ in infinitely many points. Pick $x_{n_1}$ in $B(p, 1)$, $x_{n_2}$ with $n_2 > n_1$ in $B(p, \frac{1}{2})$, and so on, by recursion. This defines a convergent subsequence of $(x_n)$ that converges to $p$. So $X'$ is relatively countably compact.
If $X'$ is relatively countably compact, and $X$ is metric, then we do get that $X'$ is relatively compact (this is due to Hausdorff, IIRC). I cannot reconstruct a proof right away, but there is one here, e.g. (but this involves Cauchy filters etc.)
It turns out that these notions are also equivalent in some topological vector spaces (spaces of the form $C_p(X)$ where $X$ is compact (Grothendieck), and weak topologicals on normed vector spaces (Eberlein-Smulian)), but these are a bit more involved, I think.
Best Answer
As the space of all right continuous functions with left limits, I believe $X$ is not necessarily a metric space, since it is not generally bounded. Therefore, the equivalence between sequential compactness and compactness does not necessarily hold for $Y$.
If you add the condition that the functions in $X$ be bounded, then then space can be expressed as a metric space, equipped with the $sup$ norm
By construction, subset $Y$ itself is a sequence of functions where $Y$ is defined as $\{f^{n}(t)\}$ and so you are correct in that $\{f^{n}(t)\}$ must have a convergent subsequence.
Note $\{f^{n}(t)\}$ itself need not converge as it is the closure of $Y$ that is compact.