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 $0 \longrightarrow T \longrightarrow M \longrightarrow F \longrightarrow 0$ 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 $\mathrm{Ext}^1(F,T)$?
(Q2) Do all elements of $\mathrm{Ext}^1(F,T)$ give symtrivial modules?
Proofs:
(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 $a \otimes b$, choose $n \geq 0$ such that $p^n a = p^n b = p^n x = 0$. Choose $u,v \in \mathbb{Z}$ with $a \equiv ux \pmod {p^n}$ and $b \equiv vx \pmod {p^n}$. Then $a \otimes b = ux \otimes b = ux \otimes vx = vx \otimes ux = b \otimes ux = b \otimes a$.
(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 $R_p \otimes T_p=T_p$ 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.
Answers to Q1 and Q2:
Q1: We disambiguate: "strongly locally cyclic" means that its localization at each prime is cyclic. "weakly locally cyclic" means that each finitely generated submodule is cyclic. $K_p/R_p$ is in the second class but not the first.
We can write $T$ as the direct sum of divisible torsion module $D$ and a strongly locally cyclic module $C$. Divisible modules are injective, so $Ext^1(F,T)=Ext^1(F,C)$.
$Ext^1(F,C)=\left(\prod_p C_p\right)/C$
Coose any nontrivial homomorphism $R \to F$, then the morphism is injective and the kernel is weakly locally cyclic, say $L$. We have a long exact sequence
$Hom(L,C) \to Hom(F,C) \to Hom(R,C) \to Ext^1(L,C) \to Ext^1(F,C) \to Ext^1(R,C)$
$Hom(R,C)=C$, $Ext^1(R,C) = 0$
$Hom(F,C) = 0$ because $F$ is $p$-divisible for all $p$ such that $C_p$ is nontrivial, so the image of any homomorphism inside $C_p$ is trivial, so the whole homomorphism is trivial.
$Ext^1(L,C) = \prod_p Ext^1(L_p,C_p)$ because $L$ and $C$ are torsion. Whenever $C_p$ is nonzero, $L_p=K_p/R_p$ so $Ext^1(L_p,C_p)=Ext^1(K_p,R_p,R/p^a) = R/p^a=C_p$. So we have a short exact sequence
$0 \to C \to \prod_p C_p \to Ext^1(F,C) \to 0$
If we're interested in classifying extensions abstractly, we just have to mod out by the action of the automorphism group.
Q2: All are symtrivial.
Indeed, let $M$ be any module whose torsion and non-torsion parts satisfy the given conditions, and choose $a \otimes b \in M \otimes M$. Choose $x \in M$ such that $a \in \langle x,t \rangle$, $b\in \langle x,t\rangle$. then $a \otimes b= (k_1 x+ t_1) \otimes (k_2 x + t_2) = k_1k_2 x\otimes x + k_1 x\otimes t_2 + k_2 t_1 \otimes x + t_1 \otimes t_2$. Then $t_1 \otimes t_2 = t_2 \otimes t_1$, so the only thing that remains to check is if $t_1\otimes x=x\otimes t_1$. Suppose $n t_1=0$, then choose $y$ such that $ny \equiv x$ mod $T$, so
$x\otimes t_1=(x-ny)\otimes t_1 =t_1 \otimes (x-ny) = t_1 \otimes x$.