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As far as (semi-)stability goes, since $U$ is not projective, it doesn't make much sense to ask on $U$. You could ask if it is relatively semi-stable, or semi-stable on the fibers. From the short exact sequence $(\star)$ you can see that its degree on any fiber is $0$ and an argument similar to the proof of Claim 2 says that any sub-line bundle would have to have degree at most $0$. (For any sub line bundle, if the induced map to $\mathscr O$ is non-trivial, then this is clear, if it is trivial, then it has to be contained in the kernel, which is also $\mathscr O$, so again clear.) So, this is actually semi-stable on every fiber. Since it contains a copy of $\mathscr O$, it is not stable.

[This part does not require a restriction on the characteristic.]

As far as (semi-)stability goes, since $U$ is not projective, it doesn't make much sense to ask on $U$. You could ask if it is relatively semi-stable, or semi-stable on the fibers. From the short exact sequence $(\star)$ you can see that its degree on any fiber is $0$ and an argument similar to the proof of Claim 2 says that any sub-line bundle would have to have degree at most $0$. (For any sub line bundle, if the induced map to $\mathscr O$ is non-trivial, then this is clear, if it is trivial, then it has to be contained in the kernel, which is also $\mathscr O$, so again clear.) So, this is actually semi-stable on every fiber. Since it contains a copy of $\mathscr O$, it is not stable.

It does not split. It's semi-stable, but not stable on every fiber.

It's semi-stable, but not stable on every fiber. Assuming that the characteristic is zero this sheaf does not split. This may be true in positive characteristic, but as Damian Rössler points out the proof below requires characteristic zero.

Proof. We may obviously assume that $f$ is smooth over $U$. Since $f$ is a non-isotrivial fibration, the associated Kodaira-Spencer map is non-zero, that is, pushing forward the dual of the original sequence, $$0 \to T_{X/\mathbb P^1} \to T_X \to f^*T_{\mathbb P^1} \to 0,\tag{$\star^\vee$}$$ gives an injective map (sometimes called the Kodaira-Spencer map) $$ \kappa: T_{\mathbb P^1}\to R^1f_* T_X. $$ Clearly, this is injective on any non-empty open set $U$ which implies that $(\star^\vee)|_U$ cannot be split and hence the Claim is proven. $\square$

Remark The fact that this sequence does not split does not mean that the sheaf in the middle cannot be the direct sum of two line bundles. Contemplate the Euler sequence of $\mathbb P^1$: $$ 0 \to \omega_{\mathbb P^1} \to \mathscr O_{\mathbb P^1}(-1) \oplus \mathscr O_{\mathbb P^1}(-1)\to \mathscr O_{\mathbb P^1} \to 0. $$

Proof. We may obviously assume that $f$ is smooth over $U$. Since $f$ is a non-isotrivial fibration, the associated Kodaira-Spencer morphism of sheaves is non-zero, that is, pushing forward the dual of the original sequence, $$0 \to T_{X/\mathbb P^1} \to T_X \to f^*T_{\mathbb P^1} \to 0,\tag{$\star^\vee$}$$ gives an injective map $$ \kappa: T_{\mathbb P^1}\to R^1f_* T_X. $$ Clearly, this is injective on any non-empty open set $U$ which implies that $(\star^\vee)|_U$ cannot be split and hence the Claim is proven. $\square$

Remark 1 The map $\kappa$ is the sheaf version of the classical Kodaira-Spencer map. That is obtained by tensoring with the residue field of a point on the target to get $T_{\mathbb P^1, t}\to H^1(X_t, T_{X_t})$ for $t\in \mathbb P^1$. That may be zero even for non-isotrivial families

Remark 2 The fact that this sequence does not split does not mean that the sheaf in the middle cannot be the direct sum of two line bundles. Contemplate the Euler sequence of $\mathbb P^1$: $$ 0 \to \omega_{\mathbb P^1} \to \mathscr O_{\mathbb P^1}(-1) \oplus \mathscr O_{\mathbb P^1}(-1)\to \mathscr O_{\mathbb P^1} \to 0. $$