I think your guess is correct and one can proceed as follows (some details are missing though). Let $V$ be a six dimensional symplectic vector space and $F$ be a rank three-vector bundle on $E$ wiht an exact sequence $$ 0 \longrightarrow G \longrightarrow V \otimes \mathcal{O}_E \longrightarrow F \longrightarrow 0,$$ where $G$ is a Lagrangian subbundle of $V \otimes \mathcal{O}_E$.

Assume furthermore that $\det(F) = \det(G^*) = \mathcal{O}_{E}(6P)$.

Let $W \subset V$ be a generic Lagrangian subspace and consider the map:

$$ \phi : G \longrightarrow V/W \otimes \mathcal{O}_{E}.$$

The genericity of $W$ implies that it is generically on E an ismorphism. Furthermore, $\phi$ is (globally) injective as $G$ is torsion free. We denote by $Z \subset E$ the subscheme corresponding to the degeneracy locus of $\phi$. Since $\det(G^*) = \mathcal{O}_{E}(6P)$, we have the linear equivalence $Z \sim 6P$.

We have an exact sequence:

$$ 0 \longrightarrow G \longrightarrow V/W \otimes \mathcal{O}_E \longrightarrow \mathcal{F} \longrightarrow 0,$$ where $\mathcal{F}$ is scheme theoretically supported on $Z$.

The vector space $W \subset V$ is generic and $E$ is a curve, so that the corank of $\phi$ is exactly $1$ on $Z$. As a consequence $\mathcal{F}|_{Z}$ is a line bundle on $Z$.

Let $Z_{red} = \{P_1, \ldots, P_l\}$ with the $P_i$ distincts. We write:

$$ \mathcal{F} = \bigoplus_{i=1}^{l} \mathcal{F}_i,$$

where $ \mathcal{F}_i$ is the restriction of $\mathcal{F}$ to the connected components of $Z$ corresponding to $P_i$.

For any subbundle $F$ of $V \otimes \mathcal{O}_E$ whose quotient is a vector bundle, we denote by $F^{\perp}$ the orthogonal to $F$ with respect to the symplectic form on $V$.

We have han exact sequence: $$ 0 \longrightarrow G^{\perp} \longrightarrow V/W^{\perp} \otimes \mathcal{O}_E \longrightarrow \mathcal{H} \longrightarrow 0,$$ where $\mathcal{H}$ is scheme theoretically supported on a subscheme of $E$ linearly equivalent to $6P$.

We similarly split $\mathcal{H}$ as $\bigoplus_{i=1}^q \mathcal{H}_i$, where the $\mathcal{H}_i$ correspond to the various connected component of the support of $\mathcal{H}$.

The bundles $G$ and $W \otimes \mathcal{O}_E$ being Lagrangian, the skew-symmetric form $\sigma : V \longrightarrow V^*$ induces isomorphisms:

$$ \sigma_{G} \ : \ G \stackrel{\sim}\longrightarrow G^{\perp} \ \textrm{and} \ \sigma_{V/W} \ : \ V/W \stackrel{\sim}\longrightarrow V/W^{\perp}$$ which are compatible with the maps:

$$ G \longrightarrow V/W \ \textrm{and} \ G^{\perp} \longrightarrow V/W^{\perp}.$$

We deduce that $\mathcal{H}$ and $\mathcal{F}$ are equal (and not just isomorphic) and that up to a reordering of the we have $\mathcal{H}_i = \mathcal{F}_i$, for all $i$.

For all $i \in \{1, \ldots, l\}$, the skew symmetric isomorphism $\sigma$ induces a skew-symmetric isomorphism:

$$\sigma_i : \mathcal{F}_i \stackrel{\sim}\longrightarrow \mathcal{F}_i,$$ which lifts to a skew-symmetric isomorphism:

$$h^0(\sigma_i) \ : \ H^0(E,\mathcal{F}_i) \stackrel{\sim}\longrightarrow H^0(E, \mathcal{F}_i).$$

The skew-symmetry of the isomorphism $h^0(\sigma_i)$ forces the dimension of the vector spaces $H^0(E,\mathcal{F}_i)$ to be even. As a consequence, of the Riemman-Roch formula on $E$, the multiplicity of $P_i$ as a connected component of $Z$ must always be even.

The generic situation (that is when $E \longrightarrow LG(3,6)$ is a generic point in a component of $Hom(E, LG(3,6))$ should correspond to the case:

$Z_{red} = \{A,B,C\}$ with $A,B,C$ distincts and $Z = \{2A,2B,2C\}$ as a subscheme of $E$.

Now I would like to deduce from this that we have a map:

$$ \mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C) \longrightarrow G$$ which is generically an isomorphism (I have a vague idea why this should be true, but I don't have a precise argument to offer, perhaps someone else will find).

If we have such a map which is generically an isomorphism, then it must be an isomorphism, owing to the relation $\det(G) = \det(\mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C))$.

We conclude that $F \simeq \mathcal{O}_E(2A) \oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$ as $G^* \simeq F$.

I think your guess is correct and one can proceed as follows (some details are missing though). Let $V$ be a six dimensional symplectic vector space and $F$ be a rank three-vector bundle on $E$ wiht an exact sequence $$ 0 \longrightarrow G \longrightarrow V \otimes \mathcal{O}_E \longrightarrow F \longrightarrow 0,$$ where $G$ is a Lagrangian subbundle of $V \otimes \mathcal{O}_E$.

Assume furthermore that $\det(F) = \det(G^*) = \mathcal{O}_{E}(6P)$.

Let $W \subset V$ be a generic Lagrangian subspace and consider the map:

$$ \phi : G \longrightarrow V/W \otimes \mathcal{O}_{E}.$$

The genericity of $W$ implies that it is generically on E an ismorphism. Furthermore, $\phi$ is (globally) injective as $G$ is torsion free. We denote by $Z \subset E$ the subscheme corresponding to the degeneracy locus of $\phi$. Since $\det(G^*) = \mathcal{O}_{E}(6P)$, we have the linear equivalence $Z \sim 6P$.

We have an exact sequence:

$$ 0 \longrightarrow G \longrightarrow V/W \otimes \mathcal{O}_E \longrightarrow \mathcal{F} \longrightarrow 0,$$ where $\mathcal{F}$ is scheme theoretically supported on $Z$.

The vector space $W \subset V$ is generic and $E$ is a curve, so that the corank of $\phi$ is exactly $1$ on $Z$. As a consequence $\mathcal{F}|_{Z}$ is a line bundle on $Z$.

Let $Z_{red} = \{P_1, \ldots, P_l\}$ with the $P_i$ distincts. We write:

$$ \mathcal{F} = \bigoplus_{i=1}^{l} \mathcal{F}_i,$$

where $ \mathcal{F}_i$ is the restriction of $\mathcal{F}$ to the connected components of $Z$ corresponding to $P_i$.

For any subbundle $F$ of $V \otimes \mathcal{O}_E$ whose quotient is a vector bundle, we denote by $F^{\perp} = (V/F)^*$.

We have han exact sequence: $$ 0 \longrightarrow G^{\perp} \longrightarrow V^*/(W^{\perp}) \otimes \mathcal{O}_E \longrightarrow \mathcal{H} \longrightarrow 0,$$ where $\mathcal{H}$ is scheme theoretically supported on a subscheme of $E$ linearly equivalent to $6P$.

We similarly split $\mathcal{H}$ as $\bigoplus_{i=1}^q \mathcal{H}_i$, where the $\mathcal{H}_i$ correspond to the various connected component of the support of $\mathcal{H}$.

The bundles $G$ and $W \otimes \mathcal{O}_E$ being Lagrangian, the skew-symmetric form $\sigma : V \longrightarrow V^*$ induces isomorphisms:

$$ \sigma_{G} \ : \ G \stackrel{\sim}\longrightarrow G^{\perp} \ \textrm{and} \ \sigma_{V/W} \ : \ V/W \stackrel{\sim}\longrightarrow V^*/(W^{\perp})$$ which are compatible with the maps:

$$ G \longrightarrow V/W \ \textrm{and} \ G^{\perp} \longrightarrow V^*/(W^{\perp}).$$

We deduce that $\mathcal{H}$ and $\mathcal{F}$ are equal and that up to a reordering of the we have $\mathcal{H}_i = \mathcal{F}_i$, for all $i$.

For all $i \in \{1, \ldots, l\}$, the skew symmetric isomorphism $\sigma$ induces a skew-symmetric isomorphism:

$$\sigma_i : \mathcal{F}_i \stackrel{\sim}\longrightarrow \mathcal{F}_i,$$ which lifts to a skew-symmetric isomorphism:

$$h^0(\sigma_i) \ : \ H^0(E,\mathcal{F}_i) \stackrel{\sim}\longrightarrow H^0(E, \mathcal{F}_i).$$

The skew-symmetry of the isomorphism $h^0(\sigma_i)$ forces the dimension of the vector spaces $H^0(E,\mathcal{F}_i)$ to be even. As a consequence, of the Riemman-Roch formula on $E$, the multiplicity of $P_i$ as a connected component of $Z$ must always be even.

The generic situation (that is when $E \longrightarrow LG(3,6)$ is a generic point in a component of $Hom(E, LG(3,6))$ should correspond to the case:

$Z_{red} = \{A,B,C\}$ with $A,B,C$ distincts and $Z = \{2A,2B,2C\}$ as a subscheme of $E$.

Now I would like to deduce from this that we have a map:

$$ \mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C) \longrightarrow G$$ which is generically an isomorphism (I have a vague idea why this should be true, but I don't have a precise argument to offer, perhaps someone else will find).

If we have such a map which is generically an isomorphism, then it must be an isomorphism, owing to the relation $\det(G) = \det(\mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C))$.

We conclude that $F \simeq \mathcal{O}_E(2A) \oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$ as $G^* \simeq F$.

In the orthogonal case (instead of the symplectic case), I think your guess is correct and one can proceed as follows. Let $V$ be a six dimensional orthogonal vector space and $F$ be a rank three-vector bundle on $E$ wiht an exact sequence $$ 0 \longrightarrow G \longrightarrow V \otimes \mathcal{O}_E \longrightarrow F \longrightarrow 0,$$ where $G$ is a maximal-isotropic subbundle of $V \otimes \mathcal{O}_E$.

Let $W \subset V$ be a generic maximal isotropic subspace and consider the map:

The genericity of $W$ implies that it is generically on E an ismorphism. Furthermore, it is (globally) injective as $G$ is torsion free. We denote by $Z \subset E$ the subscheme corresponding to the degeneracy locus of $\phi$. Since $\det(G^*) = \mathcal{O}_{E}(6P)$, we have the linear equivalence $Z \sim 6P$.

Let $x \in E$. The dimension of the vector space $\mathcal{F} \otimes \mathbb{C}(x)$ is equal to $\dim (E_x \cap W)$. The curve $E$ being connected, the parity of $\dim (E_x \cap W)$ does not depend on $x$ (this comes from the fact that the orthogonal Grassmannian $OG(3,6)$ has two connected components. Each one is defined by the parity of the dimension of the intersection of the isotropic subspaces it paramatrizes with a fixed maximal istropic subspace of $V$).

As a consequence, the parity of $\dim \mathcal{F} \otimes \mathbb{C}(x)$ does not depend on $x \in E$. Since for generic $x \in E$, we have $\mathcal{F} \otimes \mathbb{C}(x) = 0$, we deduce that:

$$ \dim \mathcal{F} \otimes \mathbb{C}(x) = 0,2,4 \ \textrm{or} \ 6.$$

The generic situation (that is a generic member of $\mathrm{Hom}(E, OG(3,V))$ should correspond to the case $Z_{red} = \{A,B,C\}$ (with $A,B,C$ distincts) and:

$$ \dim \mathcal{F} \otimes \mathbb{C}(A) = \dim \mathcal{F} \otimes \mathbb{C}(B) = \dim \mathcal{F} \otimes \mathbb{C}(C) = 2.$$ This implies that $Z = 2A+2B+2C$ as a subscheme of $E$.

Since we have two exact sequences: $$ 0 \longrightarrow G \longrightarrow W/V \otimes \mathcal{O}_E \longrightarrow \mathcal{F} \longrightarrow 0$$

and

$$0 \longrightarrow \mathcal{O}_{E}(-2A) \oplus \mathcal{O}_{E}(-2B) \oplus \mathcal{O}_{E}(-2C) \longrightarrow W/V \otimes \mathcal{O}_E \longrightarrow \mathcal{F} \longrightarrow 0$$

we have a map of coherent sheaves:

$$\psi : G \longrightarrow \mathcal{O}_{E}(-2A) \oplus \mathcal{O}_{E}(-2B) \oplus \mathcal{O}_{E}(-2C)$$

which is an isomorphism outside $Z$. As $E$ has dimension $1$ and $\det(G) = \det( \mathcal{O}_{E}(-2A) \oplus \mathcal{O}_{E}(-2B) \oplus \mathcal{O}_{E}(-2C))$, the cokernel of $\psi$ is supported on the emptyset and $\psi$ is surjective. The map $\psi$ is also injective (because $G$ is torsion free), it is an isomorphism.

We can conclude that $F \simeq \mathcal{O}_{E}(2A) \oplus \mathcal{O}_{E}(2B) \oplus \mathcal{O}_{E}(2C)$, because $F^* \sim G$ isotropic subspace).

I have the feeling that this argument could be adapted to the case $V$ is symplectic instead or orthogonal, but I don't see how to prove in the situation that the dimension of $\mathcal{F} \otimes \mathbb{C}(x)$ must always be even.

I think your guess is correct and one can proceed as follows (some details are missing though). Let $V$ be a six dimensional symplectic vector space and $F$ be a rank three-vector bundle on $E$ wiht an exact sequence $$ 0 \longrightarrow G \longrightarrow V \otimes \mathcal{O}_E \longrightarrow F \longrightarrow 0,$$ where $G$ is a Lagrangian subbundle of $V \otimes \mathcal{O}_E$.

Let $W \subset V$ be a generic Lagrangian subspace and consider the map:

The genericity of $W$ implies that it is generically on E an ismorphism. Furthermore, $\phi$ is (globally) injective as $G$ is torsion free. We denote by $Z \subset E$ the subscheme corresponding to the degeneracy locus of $\phi$. Since $\det(G^*) = \mathcal{O}_{E}(6P)$, we have the linear equivalence $Z \sim 6P$.

The vector space $W \subset V$ is generic and $E$ is a curve, so that the corank of $\phi$ is exactly $1$ on $Z$. As a consequence $\mathcal{F}|_{Z}$ is a line bundle on $Z$.

Let $Z_{red} = \{P_1, \ldots, P_l\}$ with the $P_i$ distincts. We write:

$$ \mathcal{F} = \bigoplus_{i=1}^{l} \mathcal{F}_i,$$

where $ \mathcal{F}_i$ is the restriction of $\mathcal{F}$ to the connected components of $Z$ corresponding to $P_i$.

For any subbundle $F$ of $V \otimes \mathcal{O}_E$ whose quotient is a vector bundle, we denote by $F^{\perp}$ the orthogonal to $F$ with respect to the symplectic form on $V$.

We have han exact sequence: $$ 0 \longrightarrow G^{\perp} \longrightarrow V/W^{\perp} \otimes \mathcal{O}_E \longrightarrow \mathcal{H} \longrightarrow 0,$$ where $\mathcal{H}$ is scheme theoretically supported on a subscheme of $E$ linearly equivalent to $6P$.

We similarly split $\mathcal{H}$ as $\bigoplus_{i=1}^q \mathcal{H}_i$, where the $\mathcal{H}_i$ correspond to the various connected component of the support of $\mathcal{H}$.

The bundles $G$ and $W \otimes \mathcal{O}_E$ being Lagrangian, the skew-symmetric form $\sigma : V \longrightarrow V^*$ induces isomorphisms:

$$ \sigma_{G} \ : \ G \stackrel{\sim}\longrightarrow G^{\perp} \ \textrm{and} \ \sigma_{V/W} \ : \ V/W \stackrel{\sim}\longrightarrow V/W^{\perp}$$ which are compatible with the maps:

$$ G \longrightarrow V/W \ \textrm{and} \ G^{\perp} \longrightarrow V/W^{\perp}.$$

We deduce that $\mathcal{H}$ and $\mathcal{F}$ are equal (and not just isomorphic) and that up to a reordering of the we have $\mathcal{H}_i = \mathcal{F}_i$, for all $i$.

For all $i \in \{1, \ldots, l\}$, the skew symmetric isomorphism $\sigma$ induces a skew-symmetric isomorphism:

$$\sigma_i : \mathcal{F}_i \stackrel{\sim}\longrightarrow \mathcal{F}_i,$$ which lifts to a skew-symmetric isomorphism:

$$h^0(\sigma_i) \ : \ H^0(E,\mathcal{F}_i) \stackrel{\sim}\longrightarrow H^0(E, \mathcal{F}_i).$$

The skew-symmetry of the isomorphism $h^0(\sigma_i)$ forces the dimension of the vector spaces $H^0(E,\mathcal{F}_i)$ to be even. As a consequence, of the Riemman-Roch formula on $E$, the multiplicity of $P_i$ as a connected component of $Z$ must always be even. 

The generic situation (that is when $E \longrightarrow LG(3,6)$ is a generic point in a component of $Hom(E, LG(3,6))$ should correspond to the case:

$Z_{red} = \{A,B,C\}$ with $A,B,C$ distincts and $Z = \{2A,2B,2C\}$ as a subscheme of $E$.

Now I would like to deduce from this that we have a map:

$$ \mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C) \longrightarrow G$$ which is generically an isomorphism (I have a vague idea why this should be true, but I don't have a precise argument to offer, perhaps someone else will find).

If we have such a map which is generically an isomorphism, then it must be an isomorphism, owing to the relation $\det(G) = \det(\mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C))$.

We conclude that $F \simeq \mathcal{O}_E(2A) \oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$ as $G^* \simeq F$.