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No, this is false, a vector bundle can be indecomposable for simple numerical reason, i.e., because the Chern polynomial is not a product of linear factors (think of the tangent bundle to $\mathbb P^2$); deforming it does not change the Chern polynomial.
Actually, the way the question was formulated so that the intersection has dimension 2. In this case, to get a positive answer you need the singularity to be Cohen-Macaulay.
If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours. I voted to close.
I suppose you want $k$ to be also algebraically closed, otherwise the subgroup of elements fixing $\overline k \subseteq K$ would be a proper normal subgroup. I don't have access to Lascar's paper, but from the title I would guess he takes $k = \overline{\mathbb Q}$.
I meant this: it is very hard to imagine this might be true, but if it were, it would probably be extremely difficult to prove. In any case, it seems that you are being given counterexamples.