By the dimension theorem, $\dim V=\dim N(T) +\dim R(T)$, so whenever these have nontrivial intersection (e.g. with $T(x, y) =(0, x)$) we don't have $V=N(T) +R(T) $.
A linear operator $T$ is a projection iff it is idempotent, i.e. $T^2 =T$.
Then any vector $x$ can be decomposed as $x=(x-Tx) \ +\ Tx\ \in N(T) +R(T)$, and if $x\in N(T) \cap R(T)$, then $Tx=0$ and $x=Ty$ for some $y$, so $x=Ty=TTy=Tx=0$.
It shows that in this case we indeed have $V=N(T)\oplus R(T)$, and $T$ effectively projects $a+b\mapsto b$.
Conversely, if $V=A\oplus B$ then the projection $a+b\ \mapsto b$ is clearly idempotent.
To talk about orthogonality, an inner product has to be considered on $V$, which induces an adjoint $T^*$ for every linear operator $T$, uniquely determined by the equation $\langle Tx, y\rangle =\langle x, T^*y\rangle$ (in an orthonormal basis, the matrix of $T^*$ is just the (complex conjugated) transpose of the matrix of $T$).
Now, if $T^2=T$, we will have $N(T) \perp R(T)$ iff $T^*=T$ (selfadjoint).
Supposed $T^*=T=T^2$, if $a\in N(T) $ and $b\in R(T) $, we have
$$\langle a, b\rangle =\langle a, Tb\rangle =\langle a, T^*b\rangle =\langle Ta, b\rangle = 0$$
meaning $a\perp b$.
Conversely, if $A\perp B, \ V=A\oplus B$ and $T=a+b\mapsto b$ is an orthogonal projection, then, with the decompositions $x=x_A+x_B, \ y=y_A+y_B$, we have
$$\langle Tx, y\rangle = \langle x_B, \, y_A+y_B\rangle =\langle x_B, y_B\rangle=\langle x, Ty\rangle$$
showing $T^*=T$.
Best Answer
Yes, you are indeed correct. Specifically, $(W^\perp)^\perp = \operatorname{\overline{span}} W$ (the closure of the span of $W$), for any subset $W$ of $V$, not necessarily a subspace. We therefore get $(W^\perp)^\perp = W$ if and only if $W$ is a closed subspace (recall that all finite-dimensional subspaces are closed).
As for orthogonal projection onto a subspace $W$, they too exist if and only if $W$ is closed. Recall that orthogonal projection produces the closest point in $W$ to a given point $x$. If $x \in \overline{W} \setminus W$, then there cannot be a closest point in $W$ to $x$!