Note that all equivalence relations you get by looking at equality modulo subspaces of $V$ have the special property that $v \equiv w$ and $v ' \equiv w'$ implies $v + \lambda v' \equiv w + \lambda w' $ for all $\lambda \in F$, that is, they are compatible with the vector space structure on $V$. Such an equivalence relation is also sometimes called a congruence relation. But of course, not all equivalence relations on $V$ are congruence relations and you already found an example for this.
However, if you restrict your focus on congruence relations, then you will get indeed a bijection
$$ \{W \:|\: W \text{ a subspace of } V\} \rightarrow \{\equiv \:|\: \equiv \text{ a congruence relation on } V\} $$
by mapping a subspace $W$ to equality modulo $W$. The inverse is given by taking the equivalence class of the zero vector.
I would argue as follows: such a map $T$ necessarily induces a unique well-defined map $\widetilde{T}: V/W \longrightarrow V$ such that $\widetilde{T} \circ \pi = T$. Here $\pi: V \longrightarrow V/W$ is the natural projection.
Indeed, if you denote by $[v]$ the elements of $V/W$, then you put
$$
\widetilde{T}([v]) = T(v) \ .
$$
This is well-defined, because if $[v] = [v']$, then $v - v' = w$, for some $w \in W$. Thus
$$
T(v) - T(v') = T(w) = 0 \qquad \Longleftrightarrow \qquad T(v) = T(v´) \ .
$$
So, no matter which representative you choose for $[v]$, you get the same value $T(v)$.
Hence you have a bijection $T \mapsto \widetilde{T}$ and $S \mapsto S \circ \pi$ between maps in $T \in E$ and maps $S\in\mathrm{Hom}(V/W , V)$.
Indeed, if you start with some $T \in E$, produce your $\widetilde{T}$ and then go back $\widetilde{T} \circ \pi = T$, by definition of $\widetilde{T}$.
The other way around: start with some $S: V/W \longrightarrow V$, compose with $\pi$ and get $\widetilde{S\circ \pi}$. You necessarily have $\widetilde{S\circ \pi} = S$ (exercise :-) ).
The dimension of the latter is $\mathrm{dim}(V/W) \times \mathrm{dim}(V)$.
EDIT. (Answering a question in the comments.) Let $T: V \longrightarrow U$ be a linear map. It doesn't matter if $V = U$, or not. Let $W \subset V$ be a subspace. You can always build a basis of $V$ as follows: take a basis $w_1, \dots, w_p$ of $W$ and then add the necessary vectors to get a basis of the whole space $V$: $w_1, \dots, w_p, v_{p+1}, \dots, v_n$. Take any basis $u_1, \dots, u_m$ of $U$. Then the matrix associated with $T$ in these bases has the form $(A | B)$, where the first $p$ columns $A$ are the coordinates of $Tw_1, \dots, Tw_p$ in the basis $u_1, \dots, u_m$ and the remaining $n-p$ columns $B$ are the coordinates of $Tv_{p+1}, \dots, Tv_n$. If $W \subset \mathrm{ker}T$, then the first $p$ columns are obviously zero, and your matrix has always the form $(0 | B)$. Submatrix $B$ is also the matrix of $\widetilde{T}: V/W \longrightarrow U$ in the bases $[v_{p+1}], \dots, [v_n]$ and $u_1, \dots, u_m$ of $V/W$ and $U$, respectively.
Best Answer
Regarding the first part, your proof is perfectly fine. You can shorten your proof to the following. First, note that $U/T$ is a vector space by the definition of the quotient space. On the other hand, $U/T$ is also a subset of $V/T$ over which addition and scalar multiplication have the same meaning. Because $U/T$ is a subset of $V/T$ that is itself a vector space, it is by definition a subspace of $V/T$.
Regarding the second part, here is one such bijective linear map. Define $\phi:(V/T)/(U/T) \to V/U$ by $$ \phi((x + T) + U/T)) = x + U. $$ I will leave it to you to argue that this map is well defined (i.e. gives the same result regardless of which element $x$ is used to represent $x + T$), linear, and bijective.
Alternatively, you may have encountered the fact that for any linear transformation $f:V \to W$, we have $$ \operatorname{im}(f) \cong V/\ker(f). $$ With this in mind, you could also consider the map $f:V/T \to V/U$ given by $$ f(x + T) = x + U. $$ verify that this map is well defined, that it is surjective, and that its kernel is $U/T$.