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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

$x\otimes t_1=(x-ny)\otimes t_1 =t_1 \otimes (x-ny) = t_1 \otimes x$.

Let $R$ be a Dedekind domain with field of fractions $K$. A module $M$ is symtrivial if and only if its maximal torsion-free quotient $F$ is a submodule of $K$ and its torsion submodule $T$ is a sum over each prime $p$ such that $F$ is $p$-divisible of a $p$-divisible $p$-torsion module and a $p$-power cyclic module $R/p^n$.

If $F$ and $T$ are modules as described, then we can classify extnesions $0 \to T \to M \to F \to 0$ by the group $Ext^1(F,T)$. If $F\neq 0$, this group is equal to $\prod_p R/p^n / \sum_p R/p^n$.

Proof: First we prove that such a module is symtrivial. To do this, let $a$ and $b$ be two elements and let $I$ be the submodule of $F$ generated by $a$ and $b$ modulo $T$. $I$ is a finitely generated submodule of $M$, hence it is projective, so we can find an injection $I \to M$. Hence $a$ and $b$ can be written as $x_1+t_1$, $x_2+t_2$ with $x_1,x_2 \in I$, $t_1,t_2 \in T$. We need to check:

a. $x_1 \otimes x_2 = x_2\otimes x_1$ for $x_1,x_2 \in I$.

b. $x \otimes t= t\otimes x$ for $x\in I$, $t\in T$

c. $t_1\otimes t_2 = t_2\otimes t_2$ for $t_1,t_2, \in T$.

Proof of a: $I$ is either $0$ or a fractional ideal. The case $I=0$ is trivial, so we may assume that $I$ is a fractional ideal. $I\otimes I$ is $I^2$, another fractional ideal, and the multiplication is clearly commutative.

Proof of b: $t$ is torsion, so assume $n t =0$ for some $n \in R$. For each prime $p$ dividing $n$, we know that $F$ is $p$-divisible, so $x$ is $n$-divisible up to torsion. In other words, we can write $x=ny+t'$ for $t'$ torsion. Then

$$x\otimes t= (ny+t') \otimes t = n (y\otimes t)+(t'\otimes t)= (t'\otimes t)$$

This reduces this case to case c.

Proof of c: Write $T= C+D$ for $C=\sum_p R/p^n$ and $D$ a divisible torsion module. Since $D$ is divisible, the tensor product of $D$ with any torsion module is $0$. Hence $T\otimes T = (C+D)\otimes (C+D)= C\otimes C$. But $C\otimes C = \sum_p R/p^n \otimes R/p^n = \sum_p R/p^n = C$, where the isomorphism is given by the multiplicative structure of $R/p^n$, which is clearly commutative, so $t_1\otimes t_2 = t_2 \otimes t_1$.

Next we prove that any symtrivial module has this form. Given a symtrivial $M$, we write it as an exact sequence $0 \to T \to M \to F \to 0$, with $T$ torsion and $F$ torsion-free. First we show that $F \subseteq K$. This just follows Todd Trimble's argument from the comments - tensoring with $K$ is symmetric monoidal, so preserves symtriviality, so it takes $M$ to a symtrivial vector space, which must be $0$ or $1$-dimensional. The image of the map $M \to M \otimes_R K$ is just $F$, so $F \subseteq K$.

Next we must show that $T_p=0$ if $F$ is not $p$-divisible and is a sum of a cyclic module and a $p$-divisible $p$-module otherwise. To do this we may consider the localization $M_p$, which fits into an exact sequence $0 \to T_p \to F_p \to M_p$. $M_p$ must be symtrivial. $M_p \subset R_p$. $F_p \subseteq K_p$, so $F_p=0,R_p$, or $K_p$. The middle case is the non-$p$-divisible case, and it clearly forces $T_p=0$, since $R_p$ is projective so the exact sequence splits, and we have $(R_p + T_p) \otimes (R_p + T_p) = R_p+T_p+T_p+ T_p\otimes T_p$, with the two $T_p$s switched by the symmetry, so they must both vanish.

If $F_p=0$, $T_p$ is symtrivial. We will demonstrate that if $F_p=K_p$, $T_p$ is symtrivial still. Suppose not. Find an $a,b$ in $T_p$ such that $a\otimes b \neq b \otimes a$ in $T_p \otimes T_p$, but $a \otimes b = b \otimes a$ in $M_p \otimes M_p$. The proof that $a \otimes b = b\otimes a$ must involve finitely many elements of $R_p$,. Consider the submodule $M'$ of $M_p$ generated by $T_p$ and those finitely many elements. The torsion-free quotient of $M'$ is a finitely generated submodule of $K_p$, thus is $R_p$ or $0$, thus is projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T_p\otimes T_p$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T_p\otimes T_p$, 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.

So $T_p$ is symtrivial. We must show it is a sum of a $p$-divisible $p$-torsion module and a $p$-power cyclic module. Since tensoring with $R/p$ is a symmetric monodical functor, $T_p/pT_p$ is symtrivial, so $T_p/pT_p = 0$ or $R/p$. In the first case, $T_p$ is a $p$-divisible $p$-torsion module, so we are done. So the $p$-adic completion of $T_p$ is, by Nakayama's Lemma, generated by a single element. The $p$-adic completion has the form $\hat{R}_p/p^n$ for some $n$, and $n$ must be finite because otherwise the completion morphism would be a nontrivial morphism to a $\hat{R}_p$, a torsion-free module, which is impossible. The kernel of $T_p \to \hat{T}_p$ is $p$-divisible because its elements are exactly the elements of $T_p$ that are in $p^n T_p$ for each $n$. Since it is $p$-divisible $p$-torsion, it is injective, so the exact sequence $0 \to \cap_n p^n T_p \to T_p \to R/p^n \to 0$ splits. (The last map is a surjection because every map to $R/p^n$ that projects nontrivially onto $R/p$ is surjective.)

Finally, we compute the Ext group. Again write $T= C+D$ where $C= \sum_p R/p^n$ adn $D$ a divisible torsion module. $Ext^1(F,T)= Ext^1(F,C)+Ext^1(F,D)$, but $D$ is injective so $Ext^1(F,D)=0$. and we just need to find $Ext^1(F,C)$. We will do so using the following lemma:

Lemma: Let $X$ be a torsion $R$-module that is the sum of finite $p$-torsion modules for different primes $p$, so $X = \sum_p X_p$. Let $Y$ be a torsion-free $R$-module that is $p$-divisible for each $p$ such that $X_p$ is nontrivial. Then

$$Ext^1(Y,X) = Hom(Y \otimes_R K, \prod_p X_p / \sum_p X_p )$$

Proof: We use the exact sequence $0\to \sum_p X_p \to \prod_p X_p \to \prod_p X_p / \sum_p X_p$. We obtain a long exact sequence, the relevant terms are $Hom(Y, \prod_p X_p) \to Hom(Y, \prod_p X_p / \sum_p X_p) \to Ext^1(Y,X) \to Ext^(Y,\prod_p X_p)$. We can pull out the product, replacing the first term with $\prod_p Hom( Y,X_P)$ and the second term with $Ext^1(Y, X_p)$. These modules must be $p$-divisible and must be boundedly $p$-torsion, so they vanish, and hence we get an isomorphism $Hom(Y, \prod_p X_p / \sum_p X_p) =Ext^1(Y,X)$. But $\prod_p X_p/ \sum_p X_p$ is divisible, and torsion-free, hence it is a $K$-vector space, so the Homs uniquely factor thoruhg $Y \otimes_R K$.

In our case setting $X=C$, $Y=F$, $Y \otimes_R K$ is a one-dimensional vector space if $F$ is nontrivial, so the Homs are just $\prod_p R/p^n / \sum_p R/p^n$

(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 $\mathrm{Ext}^1(T,F)$?

(Q2) Do all elements of $\mathrm{Ext}^1(T,F)$ give symtrivial modules?

Proof

(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.

(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:

(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

$x\otimes t_1=(x-ny)\otimes t_1 =t_1 \otimes (x-ny) = t_1 \otimes x$.

(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.

(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?

(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 $n$

(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 $T_p=0$.

(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.

(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 $\mathrm{Ext}^1(T,F)$?

(Q2) Do all elements of $\mathrm{Ext}^1(T,F)$ give symtrivial modules?

(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 $a \otimes b = ux \otimes b = ux \otimes vx = vx \otimes ux = b \otimes ux = b \otimes a$.

(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 $T_p=0$.