I recommend looking at my earlier post; the map I call $\alpha$ in that context is what you refer to as $E^{(V)}$. As I explain in this post, the "natural" identification between $V$ and $V^{**}$ is the map $E^{(V)}:V \to V^{**}$ given by $E^{(V)}(v) = E_v$ for $v \in V$, where $E_v \in V^{**}$ is the map from $V^*$ to the underlying field $F$ given by $E_v(f) = f(v)$, for any $f \in V^*$.
Informally, for any $v \in V$, $E_v$ describes the "natural" way in which a vector $v$ can be used to make a linear map $f:V \to F$ into an element of $F$.
Now, we also need to understand what $T^{**}$ means. First, we define $T^*$: any map $T:V \to W$ induces a dual map $T^*:W^* \to V^*$. For any $f \in W^*$, this map is defined such that $T^*(f) = f \circ T$. We can take the dual map of this dual map to get the map $T^{**}:V^{**} \to W^{**}$, which we call the "double dual" of the original map $T$. For any $\phi \in V^{**}$, this map is defined so that $T^{**}(\phi) = \phi \circ T^{*}$
Your question, then, is why it holds that $T^{**} \circ E^{(V)} = E^{(W)} \circ T$. Note that both sides of this equation describe a linear transformation, which is to say that they describe a function (which the special property of "linearity"). The two functions $T_1 = T^{**} \circ E^{(V)}$ and $T_2 = E^{(W)} \circ T$ from $V$ to $W^{**}$ are the same if for every input $v \in V$ (the common domain of these functions), we have $T_1(v) = T_2(v)$ (i.e. the two functions produce the same output). That is, we need to show that for every $v \in V$, we have
$$
(T^{**} \circ E^{(V)})(v) = (E^{(W)} \circ T)(v).
$$
Now, we're actually in the same situation as before: both sides of the equation describe an element of $W^{**}$, which is to say that both sides describe a linear transformation from $W^*$ to $F$. Again, the two functions are the same if they take every input $f \in W^*$ to the same output. That is, we need to show that for every $f \in W^{**}$,
$$
[(T^{**} \circ E^{(V)})(v)](f) = [(E^{(W)} \circ T)(v)](f).
$$
Finally, we have an equation between two numbers whose truth we need to confirm. To simplify the left-hand side, note that
\begin{align}
[(T^{**} \circ E^{(V)})(v)](f) &=
[(T^{**}(E^{(V)}(v))](f) = [T^{**}(E_v)](f)
\\ &= [E_v \circ T^*](f) = E_v(T^*(f)) = E_v(f \circ T)
\\ &= (f \circ T)(v) = f(T(v)).
\end{align}
For the right-hand side,
\begin{align}
[(E^{(W)} \circ T)(v)](f) &= [E^{(W)}(T(v))](f) = E_{T(v)}(f) = f(T(v)).
\end{align}
So, the two sides are indeed the same, which means that $[(T^{**} \circ E^{(V)})(v)]$ and $[(E^{(W)} \circ T)(v)]$ are the same element of $W^{**}$, which means that $T^{**} \circ E^{(V)} = E^{(V)} \circ T$, which was what we wanted.
