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Addendum: The examples above are (I think) correct, but I am kind of embarrasssed by how unnecessarily complicated they are. My compulsion to put the blowup of $\mathbf P^2$ in 9 points in every answer I write led me to overlook much simpler examples that fit the bill just as well. So let me add those in now. (And then I will leave this answer alone.)

For the first version, now let $X$ just be $\mathbf P^2$ blown up in a single point, and $f:X \rightarrow \mathbf P^1$ the projective bundle map. Again let $D$ be the exceptional divisor of $X \rightarrow \mathbf P^2$. Then everything in the previous answer works the same way. In this case the fibre $F$ is $\mathbf P^1$, so $D_{|F}$ is already basepoint-free, even very ample.

For the second version, where the stable base locus of $D$ is meant to have codimension 2, we can take $X$ to be the projective bundle $$X = \mathbf P_{\mathbf P^1} \left( O \oplus O(-1) \oplus O(-1) \right) $$ and our morphism is again the bundle map $X \rightarrow \mathbf P^1$.

The point is that $X$ is a compactification of the total space of the bundle $O(-1)\oplus O(-1)$ on $\mathbf P^1$. So it contains a curve $C$ which is a copy of $\mathbf P^1$ with normal bundle $O(-1) \oplus O(-1)$. I claim this curve can be flopped to give a pseudoisomorphism $X \dashrightarrow W$, and taking a very ample divisor on $W$ and its proper transform on $X$ we again get a divisor class $D$ on $X$ whose base locus is precisely $C$, but whose restriction to any fibre $F \cong \mathbf P^2$ is basepoint-free, even very ample.

These examples have the virtue of being much simpler than the first set, but also showing that this behaviour is probably very common --- you don't need special conditions like having a fibration with fibres which are abelian varieties, for example.

First, there is a small inclarity in the question with the meaning of "restriction of the base-locus". Since the base locus is in general just a closed set, of possibly arbitrary codimension, it is not clear to me whether "restriction" is just meant to mean set-theoretic intersection, or something more sophisticated.

In any case, even in the simplest situation where the base locus is a codimension 1, the answer to both questions is no.

For the first version, let $X$ be $\mathbf P^2$ blown up in 9 points which are the intersection of two smooth cubics. Let $X \rightarrow \mathbf P^1$ be the elliptic fibration, and let $D$ be one of the exceptional divisors of $X \rightarrow \mathbf P^2$. Then $mD$ is fixed for all $m \geq 0$, but for a smooth fibre $F$ the restriction $D_{|F}$ is a point on an elliptic curve, so $2D_{|F}$ is basepoint free.

For the second version, fix an ample divisor $A$ on $X$ as above. For $D$ as before and for any $n \geq 1$, the divisor $A+nD$ is big. For $n$ sufficiently large, we have $(A + nD) \cdot D = A \cdot D -n <0$, so for such $n$ and sufficiently large $m$, the base locus of $m(A+nD)$ is again $D$. On the other hand, for any $n \geq 1$ the restriction of $A+nD$ to $F$ is basepoint free.

Update: Jérémy asks further for an example where the stable base locus of $D$ is codimension 2. This is a little trickier, but I think one can cook it up starting from the example constructed by Totaro in Theorem 6.1 of the paper:

Totaro, Burt, Hilbert’s 14th problem over finite fields and a conjecture on the cone of curves, Compos. Math. 144, No. 5, 1176-1198 (2008). ZBL1157.13006.

In this example we have a fibration $W \rightarrow \mathbf P^3$ whose general fibres $F$ is an abelian surface, and moreover there is a pseudo-isomorphism $X \dashrightarrow \mathbf W$ given by the inverse flip of 45 disjoint curves on the variety $X$, which is the blowup of $\mathbf P^5$ in a certain arrangement of 9 points.

So start with a very ample divisor $H$ on $X$ say. Then its proper transform $\widetilde{H}$ on $W$ will be movable, with stable base locus $Bs(\widetilde{H})$ consisting of exactly the indeterminacy locus of $W \dashrightarrow X$. This is a union of 45 copies of $\mathbf P^3$ on $W$, each of them a section of $W \rightarrow \mathbf P^3$. So the intersection of $Bs(\widetilde{H})$ with a general fibre $F$ consists of 45 points on $F$.

On the the other hand the restriction $\widetilde{H}_{|F}$ is an effective divisor on the abelian surface $F$, so it is semi-ample.

Now I missed one condition of the original question, namely that the base $Y$ should be a curve. So take $Y \subset \mathbf P^3$ to be a general line, and restrict the fibration $W \rightarrow \mathbf P^3$ to $Y$, to get a fibration $V \rightarrow Y$ say. Then taking the divisor $\widetilde{H}_{|V}$ the same argument as before will work.

First, there is a small inclarity in the question with the meaning of "restriction of the base-locus". Since the base locus is in general just a closed set, of possibly arbitrary codimension, it is not clear to me whether "restriction" is just meant to mean set-theoretic intersection, or something more sophisticated.

In any case, even in the simplest situation where the base locus is a codimension 1, the answer to both questions is no.

For the first version, let $X$ be $\mathbf P^2$ blown up in 9 points which are the intersection of two smooth cubics. Let $X \rightarrow \mathbf P^1$ be the elliptic fibration, and let $D$ be one of the exceptional divisors of $X \rightarrow \mathbf P^2$. Then $mD$ is fixed for all $m \geq 0$, but for a smooth fibre $F$ the restriction $D_{|F}$ is a point on an elliptic curve, so $2D_{|F}$ is basepoint free.

For the second version, fix an ample divisor $A$ on $X$ as above. For $D$ as before and for any $n \geq 1$, the divisor $A+nD$ is big. For $n$ sufficiently large, we have $(A + nD) \cdot D = A \cdot D -n <0$, so for such $n$ and sufficiently large $m$, the base locus of $m(A+nD)$ is again $D$. On the other hand, for any $n \geq 1$ the restriction of $A+nD$ to $F$ is basepoint free.

Update: Jérémy asks further for an example where the stable base locus of $D$ is codimension 2. This is a little trickier, but I think one can cook it up starting from the example constructed by Totaro in Theorem 6.1 of the paper:

Totaro, Burt, Hilbert’s 14th problem over finite fields and a conjecture on the cone of curves, Compos. Math. 144, No. 5, 1176-1198 (2008). ZBL1157.13006.

In this example we have a fibration $W \rightarrow \mathbf P^3$ whose general fibre $F$ is an abelian surface. The key point is that there is a pseudo-isomorphism $X \dashrightarrow \mathbf W$ given by the inverse flip of 45 disjoint curves on another variety $X$, which is the blowup of $\mathbf P^5$ in a certain arrangement of 9 points.

So start with a very ample divisor $H$ on $X$ say. Then its proper transform $\widetilde{H}$ on $W$ will be movable, with stable base locus $Bs(\widetilde{H})$ consisting of exactly the indeterminacy locus of $W \dashrightarrow X$. This is a union of 45 copies of $\mathbf P^3$ on $W$, each of them a section of $W \rightarrow \mathbf P^3$. So the intersection of $Bs(\widetilde{H})$ with a general fibre $F$ consists of 45 points on $F$.

On the the other hand the restriction $\widetilde{H}_{|F}$ is an effective divisor on the abelian surface $F$, so it is semi-ample.

Now I missed one condition of the original question, namely that the base $Y$ should be a curve. So take $Y \subset \mathbf P^3$ to be a general line, and restrict the fibration $W \rightarrow \mathbf P^3$ to $Y$, to get a fibration $V \rightarrow Y$ say. Then taking the divisor $\widetilde{H}_{|V}$ the same argument as before will work.

Let me expand on my comments at a bit more length.

First, there is a small inclarity in the question with the meaning of "restriction of the base-locus". Since the base locus is in general just a closed set, of possibly arbitrary codimension, it is not clear to me whether "restriction" is just meant to mean set-theoretic intersection, or something more sophisticated.

In any case, even in the simplest situation where the base locus is a codimension 1, the answer to both questions is no.

For the first version, let $X$ be $\mathbf P^2$ blown up in 9 points which are the intersection of two smooth cubics. Let $X \rightarrow \mathbf P^1$ be the elliptic fibration, and let $D$ be one of the exceptional divisors of $X \rightarrow \mathbf P^2$. Then $mD$ is fixed for all $m \geq 0$, but for a smooth fibre $F$ the restriction $D_{|F}$ is a point on an elliptic curve, so $2D_{|F}$ is basepoint free.

For the second version, fix an ample divisor $A$ on $X$ as above. For $D$ as before and for any $n \geq 1$, the divisor $A+nD$ is big. For $n$ sufficiently large, we have $(A + nD) \cdot D = A \cdot D -n <0$, so for such $n$ and sufficiently large $m$, the base locus of $m(A+nD)$ is again $D$. On the other hand, for any $n \geq 1$ the restriction of $A+nD$ to $F$ is basepoint free.

First, there is a small inclarity in the question with the meaning of "restriction of the base-locus". Since the base locus is in general just a closed set, of possibly arbitrary codimension, it is not clear to me whether "restriction" is just meant to mean set-theoretic intersection, or something more sophisticated.

In any case, even in the simplest situation where the base locus is a codimension 1, the answer to both questions is no.

For the first version, let $X$ be $\mathbf P^2$ blown up in 9 points which are the intersection of two smooth cubics. Let $X \rightarrow \mathbf P^1$ be the elliptic fibration, and let $D$ be one of the exceptional divisors of $X \rightarrow \mathbf P^2$. Then $mD$ is fixed for all $m \geq 0$, but for a smooth fibre $F$ the restriction $D_{|F}$ is a point on an elliptic curve, so $2D_{|F}$ is basepoint free.

For the second version, fix an ample divisor $A$ on $X$ as above. For $D$ as before and for any $n \geq 1$, the divisor $A+nD$ is big. For $n$ sufficiently large, we have $(A + nD) \cdot D = A \cdot D -n <0$, so for such $n$ and sufficiently large $m$, the base locus of $m(A+nD)$ is again $D$. On the other hand, for any $n \geq 1$ the restriction of $A+nD$ to $F$ is basepoint free.

Update: Jérémy asks further for an example where the stable base locus of $D$ is codimension 2. This is a little trickier, but I think one can cook it up starting from the example constructed by Totaro in Theorem 6.1 of the paper:

Totaro, Burt, Hilbert’s 14th problem over finite fields and a conjecture on the cone of curves, Compos. Math. 144, No. 5, 1176-1198 (2008). ZBL1157.13006.

In this example we have a fibration $W \rightarrow \mathbf P^3$ whose general fibres $F$ is an abelian surface, and moreover there is a pseudo-isomorphism $X \dashrightarrow \mathbf W$ given by the inverse flip of 45 disjoint curves on the variety $X$, which is the blowup of $\mathbf P^5$ in a certain arrangement of 9 points.

So start with a very ample divisor $H$ on $X$ say. Then its proper transform $\widetilde{H}$ on $W$ will be movable, with stable base locus $Bs(\widetilde{H})$ consisting of exactly the indeterminacy locus of $W \dashrightarrow X$. This is a union of 45 copies of $\mathbf P^3$ on $W$, each of them a section of $W \rightarrow \mathbf P^3$. So the intersection of $Bs(\widetilde{H})$ with a general fibre $F$ consists of 45 points on $F$.

On the the other hand the restriction $\widetilde{H}_{|F}$ is an effective divisor on the abelian surface $F$, so it is semi-ample.

Now I missed one condition of the original question, namely that the base $Y$ should be a curve. So take $Y \subset \mathbf P^3$ to be a general line, and restrict the fibration $W \rightarrow \mathbf P^3$ to $Y$, to get a fibration $V \rightarrow Y$ say. Then taking the divisor $\widetilde{H}_{|V}$ the same argument as before will work.