If $R$ is an equivalence relation on a set $X$, we naturally get the quotient set $X/R$ whose elements are the equivalence classes under $R$. Almost all constructions of "quotient spaces" (of vector spaces, abelian groups, whatever) are then as follows: define the structure on $X/R$ so as to make the canonical map $X\rightarrow X/R$ (which takes every element to its equivalence class) a morphism.
An example which you should have seen: if $G$ is a group and $N$ a normal subgroup, define an equivalence relation $R$ on $G$ by $xRy$ iff there is $n\in N$ with $y=nx$. The equivalence classes are exactly the cosets, and the quotient set is denoted $G/N$ and is the quotient group when its group structure is defined so as to make $G\rightarrow G/N$ a homomorphism.
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
For any $\;f\in \mathcal C(a,b]\;$, define the scalar $\;k(f,a):= f(a)\;$ , and then define
$$\phi: \mathcal C[a,b]\to W\;,\;\;\phi(f):= f(x)-k(f,a)$$
Check stuff...and there you go.