We can describe a symtrivial module over a Dedekind domain $R$ with field of fractions $K$ as follows:
(1) Torsion-free symtrivial modules are submodules of $K$.
(2) The torsion part of a symtrivial module is symtrivial
(3) A torsion module is symtrivial if and only if it is a direct sum of symtrivial $p$-power-torsion modules
(4) A $p$-power-torsion module $M$ is symtrivial if and only if it satisfies either $M/pM=0$ or $M/pM=R/pR$,.
(5) If a $p$-power-torsion module $M$ satisfies $M/pM=R/pR$, then it is the direct sum of a $p$-divisible $p$-torsion module and a cyclic module. (Obviously if $M/pM=0$ then it is a $p$-divisible $p$-torsion module.)
(6) If $0 => T => M => F => 0$ is an exact sequence, with $T$ torsion, $M$ symtrivial, and $F$ torsion-free, then for each prime $p$, either $F$ is $p$-divisible or $T$ has no $p$-power-torsion.
(7) If $T$ is torsion and symtrivial, $F$ is torsion-free and symtrivial, and for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-power-torsion, then $T \oplus F$ is symtrivial.
If none of my proofs are mistaken, the remaining questions are
(Q1) For $T$ torsion symtrivial, $F$ torsion-free symtrivial, such that for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion, what is $Ext^1(T,F)$?
(Q2) Do all elements of $Ext^1(T,F)$ give symtrivial modules?
Proof
(1) This is immediate, by Todd Trimble's argument.
(2) Indeed, take a module $M$ with torsion submodule $T$ such that $a \otimes b \neq b \otimes a$ in $T\otimes T$, but $a\otimes b = b \otimes a$ in $M\otimes M$. The equality $a \otimes b = b\otimes a$ must be the consequence of finitely many relations, involving finitely many elements. Consider the submodule $M'$ generated by the whole torsion module and those finitely many elements. The torsion-free quotient of $M'$ is a finitely generated submodule of $K$, thus is a fractional ideal, thus projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T\otimes T$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T\otimes T$, they do not equal each other in $M'\otimes M'$ - but a complete set of relations implying that they do are relations of $M' \otimes M'$, a contradiction.
(3) This is immediate from things noted in the original question.
(4) Only if follows from the fact that quotients of symtrivial modules are symtrivial. For if, first note that if $M/pM=0$, then $M$ is $p$-divisible and $p$-torsion, so its tensor product with itself is trivial. If $M/pM=R/pR$, let $x$ be a lift of $1 \in R/pR = M/pM$ to $M$. For any $a \otimes b$, $a$ is $p^n$-torsion for some $n$
(5) Let $\pi$ be a uniformizer of $R_p$. This defines a surjection $p^{n-1}M/p^nM \to p^n M/p^{n+1}M$, so either $p^nM/p^{n+1}M$ is eventually $0$ or we get a nontrivial morphism $M \to \lim_n M/p^n M = \lim_n R/p^nR=\hat{R}_p$ which is torsion-free, so it's eventually $0$. Then we have a surjection $M \to M/p^n M= R/p^nR$ whose kernel is $p$-divisible. Since $p$-divisible $p$-torsion groups are divisible, they are injective, so the exact sequence splits and we get a direct sum.
(6) Localize the exact sequence at $p$. Then $F_p$ is a submodule of $K_p$, so either $R_p$ or $K_p$. In the second case $F$ is $p$-divisible so take the first case. $R_p$ is projective so the exact sequence splits into a direct sum, so $R_p \otimes T_p = 0$ by one of the notes in the question, but $R_p \otimes T_p=0$ so $T_p=0$.
(7) It suffices to prove that $T \otimes F=0$, by one of the notes in the question. But we can divide $T$ into $p$-power-torsion modules, and a $p$-divisble module tensor a $p$-power-torsion module is trivial.