Let $f: S\to S$ be a homeomorphism of a compact connected surface (possibly with boundary).
Theorem.
If $\chi(S)<0$ then the mapping torus $M=M_f$ of $f$ is a Seifert manifold if and only if the mapping class of $f$ is periodic, i.e., $f$ is isotopic to a periodic homeomorphism.
If $\chi(S)\ge 0$ then $M$ is Seifert unless $S$ is the torus and eigenvalues of the action of $f$ on $H_1(S, {\mathbb R})$ are not roots of unity.
Proof. I will prove only Part 1 since the proof is already way too long. Recall that the surface $S$ admits Thurston-Nielsen decomposition with respect to $f$, i.e., a finite collection (possibly empty) of simple loops $L_i\subset S$ which are pairwise disjoint, pairwise non-isotopic and not isotopic to the boundary of $S$ (if there is any), such that:
a. $f$ preserves the multiloop $L=\cup_i L_i$. In particular, there exists $N$ such that $f^N$ preserves
each loop $L_i$ and each component $S_j$ of $S\setminus L$.
b. The homeomorphism $f_j^N= f^N|S_j$ is homotopic to a homeomorphism $g_j: S_j\to S_j$ which is either periodic (by taking larger $N$ we can assume that such $g_j$ is the identity) or is pseudo-Anosov.
A homeomorphism $f$ is called reducible if $L$ is nonempty.
A very nice proof of this fundamental theorem can be found for instance in the book by Casson and Bleiler "Homeomorphisms of surfaces after Nielsen and Thurston."
Since a manifold is Seifert if and only if it has finite cover which is Seifert, we can assume that $S$ is oriented and $f=f^N$ (replacing $f$ with $f^N$ amounts to passing to a finite cover of $M$).
We now define tori $T_i$ which are mapping tori of the restrictions $f|L_i$. Then, by construction, the manifold $M=M_f$ is the union of submanifolds with boundary $M_j=M_{f|S_j}$, which meet along the tori $T_i$. As it was observed in the other answer, if $f|S_j$ is pseudo-Anosov, then $M_j$ is hyperbolic and, hence, $M$ cannot be a Seifert manifold in this case. (For instance, this can be seen from the fact that $\pi_1(M_j)$ has trivial center, while $\pi_1$ of a Seifert manifold always has nontrivial center, after passing to a finite cover, unless this cover is the 3-sphere.)
It remains to analyze the case when each $f|S_j$ is isotopic to the identity. Let $A_i=\eta(L_i)$ denote a small annular neighborhood of $L_i$ in $S$. Let $S_j'$ denote the complement in $S_j$ to all the annuli $A_j$ which it meets. By isotopying $f$ further, we can assume that $f_j=f|S_j'$ is the identity while the restriction $f|A_i$ is an iterated Dehn twist $D^{n_i}$ along the loop $L_i$. Then the mapping torus of $f_j$ is the product
$S_j'\times S^1$. If some $n_i$ equals zero, then $f|A_i=Id$ and we can eliminate the loop $L_i$ from $L$ without changing topology of $M$ and periodicity (or lack of thereof) of $f$.
Each annulus $A_i$ has two boundary circles which I will denote $A_i^+, A_i^-$; accordingly, each manifold $M_{f|A_i}\cong A_i\times S^1$ has two boundary tori $T_{i}^+, T_i^-$ which are the mapping tori of $f|A_i^\pm$. Each loop $A_i^\pm$ is a boundary loop of a unique component of
$$
\bigcup_j S_j'
$$
which I will denote $S_i^+$ and $S_i^-$ accordingly. (Note that it might happen that $S_i^+=S_i^-$, for instance, if $L$ is a single loop which does not separate $S$.) Accordingly, the (mapping) torus $T_{i}^\pm=M_{f|A_i^\pm}$ is a boundary component of the product manifold $M_i^\pm=M_{f|S_i^\pm}$, the mapping torus of the homeomorphism $f$ restricted to $S_i^+$ or $S_i^-$. The manifold $M_i\pm$ has canonical product structure (since $f$ restricts to the identity on $S_i^\pm$). Therefore, for each torus $T_{i}^+$ we obtain a canonical system of generators of the homology group: "Horizontal'' loop $a_{i}^+$ corresponding to the loop $A_i^+\subset S$ and the "vertical'' loop $b_i^+$ corresponding to the $S^1$-factor of the decomposition $M_i^+=S_i^+\times S^1$.
The same applies to the torus $T_{i}^-$, where we switch all pluses to minuses.
Now, consider these loops $a_i^\pm, b_i^\pm$ in the product $M_{f|A_i}=T^2\times [0,1]$. The loops $a_i^+, a_i^-$ are, of course, isotopic to each other in this product manifold, since they correspond to isotopic curves $A_i^+, A_i^-$ on the surface $S$. I will denote by $[a_i]$ their common homology class in this product manifold.
However, this is not the case for for the loops $b^+_i, b^-_i$: From the definition of the Dehn twist we obtain that
$$
[b_i^-]= [b_i^+]\pm n_i [a_i]
$$
the plus or minus depend on which boundary circle of $A_i$ we marked with $+$ and which with $-$; the number $n_i$ here is the power of the Dehn twist that we use.
The bottom line is that when under the gluing $M_i^+$ and $M_i^-$ along $A_i\times S^1$, the fibers of the (Seifert) fibration of $M_i^+$ do not match (up to isotopy) fibers of the Seifert fibration of $M_i^-$.
It is important to note here that the manifolds $M_i^\pm$ here admit unique, up to isotopy, Seifert fibrations, namely, the ones coming from their product decompositions $S_i^\pm \times S^1$, since each component of the surface $S\setminus L$ has negative Euler characteristic. (Here we are using the fact that $S$ is not the torus: cutting torus along a loop results in the annulus $A$ and the product $A\times S^1$ admits infinitely many, up to isotopy, circle fibrations.) This uniqueness theorem should be in Hempel's book on 3-manifolds and in Orlik's book on Seifert manifolds. In particular, up to isotopy, we can talk about the Seifert fibration of the manifold $M_i^\pm$.
The product regions $A_i\times S^1$ of course, admit infinitely many Seifert fibrations; to remedy this, we adjoin each $A_i\times S^1$ to the product manifold $S_i^+\times S^1$. As the result, $M$ is obtained by gluing product manifolds $M_j\cong S_j\times S^1$ along their boundary tori in such a fashion that all the gluing maps do not preserve the (up to isotopy) fibers of the circle fibrations of the manifolds $M_j$.
Now, we can finish the proof of Part 1 of the theorem with the following lemma which, I remember seeing in the book by Jaco and Shalen "Seifert fibered spaces in 3-manifolds", Memoirs of Amer. Math. Soc. 220 (1979).
Lemma. Suppose that $M$ is a 3-dimensional manifold obtained by gluing oriented Seifert manifolds $M_j$ along their incompressible boundary tori, where each $M_j$ admits a unique Seifert fibration. Then $M$ is Seifert if and only if all gluing maps preserve (up to isotopy) fibers of the Seifert fibrations.
Proof. I will skip the proof of one direction of this lemma since it is not needed and assume that one of the gluing maps does not preserve circle fibers. Let $T\subset M$ denote the incompressible torus corresponding to this gluing. Suppose, that $M$ is Seifert fibered; by looking at its fundamental group, it is clear that the base of this fibration has to be of hyperbolic type (i.e., it is a hyperbolic orbifold). It is a standard fact of the theory of Seifert manifolds (I am sure, it is in Jaco and Shalen) that every incompressible torus in such a fibration is "vertical'', i.e., isotopic to a torus foliated by Seifert fibers. To see the algebraic side of this statement, consider the short exact sequence of fundamental groups induced by Seifert fibration:
$$
1\to {\mathbb Z}\to \pi_1(M)\to B=\pi_1(O)\to 1,
$$
where $O$ is the base-orbifold of the fibration and the fundamental group of $O$ is understood in the orbifold sense. Since the torus $T$ is incompressible, its fundamental group yields a subgroup $Z^2$ of $\pi_1(M)$. Projection of this group to $B$ has to be abelian, and, by hyperbolicity assumption on $O$, infinite cyclic. Therefore, the intersection of $Z^2$ with the normal subgroup ${\mathbb Z}$ of $\pi_1(M)$ is a free (abelian) factor of $Z^2$. In particular, $T$ admits a foliation by circles where each fiber is homotopic to the generic fiber of the Seifert fibration of $M$. By working more, one promotes this to an isotopy of $T^2$.
Instead of isotopying the torus we can isotope Seifert fibration itself. Therefore, splitting $M$ along $T$ results in one or two Seifert manifolds and the gluing map preserves Seifert fibrations. Continuing inductively with respect to all the boundary tori of the manifolds $M_i$, we obtain that each $M_i$ admits a Seifert fibration and gluing maps preserve these fibrations. Since Seifert fibration of each $M_i$ was unique (up to isotopy), we are done. QED
Answer is 'always', provided that you treat your map $f$ up to homotopy and assume homotopy finiteness of the fiber. There are several cases to consider, I will deal with the case when $M$ has contractible universal cover (if you are interested in the case when it is not the case, I will write the details, the problem is solved by appealing to Kneser's theorem). Let $F$ be a homotopy fiber of your fibration. One has to assume that it is a finite cell complex, otherwise one can easily find a counter example by taking an aspherical 3-manifold whose fundamental group maps onto $Z$ with the infinitely generated kernel. (First, find a 2-dimensional example of this situation and then multiply by the circle.)
Now, by the long exact sequence of fibration you obtain that the finitely generated group $N =\pi_1(F)$ is normal in $G=\pi_1(M)$ and the quotient is the infinite cyclic group $Z$. Since Perelman proved the Poincare conjecture for our benefit, the manifold $M$ is also irreducible and, hence, you can apply a theorem which you can find somewhere in Hempel's book "3-manifolds" which states that in this case $N$ is a surface group and $M$ indeed is a fiber bundle over the circle so that the projection map of the bundle is homotopic to $f$.
For the second part. Suppose that you have a fiber bundle $f: M\to S^1$. Take $S$ to be the preimage of a point. Then, by local triviality of $f$, the complement of an open bicolor of $S$ is the product of $S$ and the unit interval. The manifold $M$ is obtained by identifying its boundary surfaces by a homeomorphism $h$, which we can regard as an automorphism of $S$. Now, it is clear that $M$ is the mapping torus of $h$. QED
Best Answer
Besides the invariants suggested in the comments, for a Nil-manifold $M_\phi$ defined by $\phi \in \text{SL}_2(\mathbb Z)$ such that $\text{trace}(\phi)=\pm 2$, the conjugacy classes of $\phi$ and $\phi^{-1}$ in the group $\text{SL}_2(\mathbb Z)$ do indeed form a topological invariant of $M_\phi$ (unlike the situation in the link you provided). Since the conjugacy classes of $\{\phi_{\pm k}\}$ and of $\{\theta_{\pm k}\}$ are all different, the mapping toruses of $M_k$ and $N_k$ are all in different homeomorphism classes.
Here's a sketch of the proof.
First, in any 3-dimensional Nil-manifold $M$, the torus fibers of all fibrations of $M$ over $S^1$ form just a single isotopy class of embedded toruses in $M$ (this is proved by arguments of 3-manifold topology; this is the part which is quite different from the link you provided).
Next, fixing one torus fiber, and choosing the "direction" transverse to the torus for the fibration over the circle, the monodromy map of the fibration falls into one of an inverse pair of mapping classes of the fiber (this is standard for all fibrations over a circle).
Finally, choosing any homeomorphism from the standard torus $T^2$ to a torus fiber, the resulting conjugacy class in $\text{SL}_2(\mathbb Z)$ falls into one of two inverse conjugacy classes, independent of all choices.