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
Best Answer
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