2 added 133 characters in body

[EDIT: as Sasha points out, this does not answer the question. Please see it as a comment explaining why $X$ should be proper!]

The answer is no in general if $X$ is not proper: take $X=\mathrm{Spec}\,\mathbb{C}[x]$, $T=\mathrm{Spec}\,\mathbb{C}[t]$, and $F=$ the structure sheaf of $Z=\mathrm{Spec}\,(\mathbb{C}[t,x]/(1-tx))$. Then $T'$ is just the origin.

Variant: if instead you take $T=\mathrm{Spec}\,\mathbb{C}[t,u]$ and $Z=\mathrm{Spec}\,(\mathbb{C}[t,u,x]/(u,1-tx))$ (i.e. the same $Z$ as before, but embedded in 3-space), then $T'$ is the union of the origin and the complement of the $t$-axis, hence not locally closed (but still constructible).

Of course the point here is that $Z$ "goes to infinity" at the origin. I don't have a counterexample where $X$ is proper, but the main problem then is "taking the dual in the fibers", as in Sasha's comment above.

1

The answer is no in general if $X$ is not proper: take $X=\mathrm{Spec}\,\mathbb{C}[x]$, $T=\mathrm{Spec}\,\mathbb{C}[t]$, and $F=$ the structure sheaf of $Z=\mathrm{Spec}\,(\mathbb{C}[t,x]/(1-tx))$. Then $T'$ is just the origin.

Variant: if instead you take $T=\mathrm{Spec}\,\mathbb{C}[t,u]$ and $Z=\mathrm{Spec}\,(\mathbb{C}[t,u,x]/(u,1-tx))$ (i.e. the same $Z$ as before, but embedded in 3-space), then $T'$ is the union of the origin and the complement of the $t$-axis, hence not locally closed (but still constructible).

Of course the point here is that $Z$ "goes to infinity" at the origin. I don't have a counterexample where $X$ is proper, but the main problem then is "taking the dual in the fibers", as in Sasha's comment above.