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Apparently I didn't read the question correctly and so this is not an answer....

This is certainly not true as stated. In other words, why would be there any lines on $V$?

Example 1 Let $V$ be a hypersurface that does not contain a line. For instance, every general surface of degree at least $4$ in $\mathbb P^3$ is such, because they have Picard number $1$ and hence cannot have any non-trivial curves on them. Any line bundle on such a $V$ is linearly trivial, because the condition is satisfied vacuously.

OK, so let's assume that $V$ contains lines.

Example 2 Let $V$ be an irrational scroll and consider the globally generated but not ample line bundle that induces the morphism that collapses the lines. This is linearly trivial because the only lines are the ones that the morphism collapses.

OK, so let's assume that $V$ is connected by (chains of) lines. However, that probably implies that $V$ is linear, so you don't gain anything.

In any case, you need a new formulation for this.

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This is certainly not true as stated. In other words, why would be there any lines on $V$?

Example 1 Let $V$ be a hypersurface that does not contain a line. For instance, every general surface of degree at least $4$ in $\mathbb P^3$ is such, because they have Picard number $1$ and hence cannot have any non-trivial curves on them. Any line bundle on such a $V$ is linearly trivial, because the condition is satisfied vacuously.

OK, so let's assume that $V$ contains lines.

Example 2 Let $V$ be an irrational scroll and consider the globally generated but not ample line bundle that induces the morphism that collapses the lines. This is linearly trivial because the only lines are the ones that the morphism collapses.

OK, so let's assume that $V$ is connected by (chains of) lines. However, that probably implies that $V$ is linear, so you don't gain anything.

In any case, you need a new formulation for this.