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
The calculation $\pi_1(V) \approx G * \mathbb{Z}$ should be pretty easy, using that $N$ is contractible: just decompose $V$ as the union of the image in $M_f$ of $N \times [0,1]$ (which is homeomorphic to $N \times S^1$) and the image in $M_f$ of $X \times ((2/3,1] \cup [0,1/3))$ (which is homeomorphic to $X \times (2/3,4/3)$).
In many (if not most) applications of the Seifert - Van Kampen theorem, besides just knowing the fundamental groups $\pi_1(U)$, $\pi_1(V)$, $\pi_1(U \cap V)$, you also need to know the homomorphisms induced by the inclusion maps
$$i_U : U \cap V \to U \quad\text{and}\quad i_V : U \cap V \to V
$$
In this case, this means that you need formulas for the two induced homomorphisms
$$G * G \approx \pi_1(U \cap V) \xrightarrow{(i_U)_*} \pi_1(U) \approx G
$$
$$G * G \approx \pi_1(U \cap V) \xrightarrow{(i_V)_*} \pi_1(V) \approx G * \mathbb{Z}
$$
You'll have to convince yourself that on one free factor of $G*G$ the restriction of $(i_U)_*$ is the identity, and on the other free factor the restriction is $f_*$.
Also, the restriction of $(i_V)_*$ to both free factors of $G*G$ is just the standard injection $G \hookrightarrow G*Z$.
With that, you should be able to complete the verification that $\pi_1(M_f,m_0) \approx \mathbb{Z} \ltimes_{f_{\ast}} G$.
Best Answer
Let $F$ be a path-connected topological space and $\phi : F \to F$ a homeomorphism. Denote the mapping torus of $\phi$ by $M$ so that we have a fiber bundle $F \to M \to S^1$. Explicitly, $M = F\times[0, 1]/\sim$ where $(x, 0) \sim (\phi(x), 1)$. From the long exact sequence in homotopy, there is a short exact sequence $$0 \to \pi_1(F) \to \pi_1(M) \to \pi_1(S^1) \to 0.$$ The covering space of $M$ corresponding to the image of $\pi_1(F)$ in $\pi_1(M)$ is homeomorphic to $F\times\mathbb{R}$. This follows from the fact that $M$ can be constructed as the quotient of $F\times\mathbb{R}$ by the $\mathbb{Z}$-action generated by $(x, t) \mapsto (\phi(x), t + 1)$. In particular, the universal cover of $M$ is homeomorphic to $\widetilde{F}\times\mathbb{R}$ where $\widetilde{F}$ is the universal cover of $F$.
Suppose now that $F$ is a closed $(n-1)$-manifold, so that $M$ is a closed $n$-manifold. If $\pi_1(M) \cong \mathbb{Z}$, then $\pi_1(F) = 0$ by the short exact sequence above. As suggested by Moishe Kohan, let $N = M \# X$ where $X$ is a closed simply connected $n$-manifold, then $\widetilde{N}$, the universal cover of $N$, is homeomorphic to the connected sum of $\widetilde{F}\times\mathbb{R} = F\times\mathbb{R}$ and countably many copies of $X$. If $H_k(X; \mathbb{Z}) \neq 0$ for some $0 < k < n$, then it follows from Mayer-Vietoris that $H_k(\widetilde{N}; \mathbb{Z}) \cong H_k(\widetilde{F}; \mathbb{Z})\oplus\bigoplus_{i\in\mathbb{Z}}H_k(X)$; in particular, $H_k(\widetilde{N}; \mathbb{Z})$ is not finitely generated. As $\pi_1(N) \cong \pi_1(M) \cong \mathbb{Z}$, if $N$ were a mapping torus, the fiber $F'$ would be a closed simply connected manifold, so $\widetilde{N}$ would be homeomorphic to $\widetilde{F'}\times\mathbb{R} = F'\times\mathbb{R}$ which has finitely generated homology. Therefore $N$ is not a mapping torus. Note, if $H_k(X; \mathbb{Z}) = 0$ for all $0 < k < n$, then $X$ is a simply connected homology sphere and hence homeomorphic to $S^n$, in which case $N$ is homeomorphic to $M$ (which is a mapping torus).
I believe that the hypothesis $\pi_1(M) \cong \mathbb{Z}$ is unnecessary, but the proof above only extends to the case where $\pi_1(M)$ is an extension of $\mathbb{Z}$ by a finite group, i.e. $F$ has finite fundamental group.