Confirm my understanding of adjoints adjoints seem REALLY important and useful so I don't want to move onto the next topic without really understanding them; I have too many a times moved on and been lost because I don't have the foundation I should.
Anyways, here goes my explanation of what I think the adjoint is. It's probably really wrong but that's what you guys are here for :):
We have a vector $v \in V$ where $V$ is an inner product space. By the Riesz Representation theorem, we can ALWAYS find a linear functional $T: V \to R$ (a transformation that takes a vector and spits out a scalar) such that $T_w(v) = \langle v,w \rangle$. 
The Riesz Representation theorem also stipulates that $w$ is unique.
This is where my first few questions arise: 


*

*Why is $w$ unique? Is it unique in that $\langle v,w\rangle$ only equals some value $x$ for a specific $w$ i.e. the linear functional is one-to-one and $x=\langle v,w\rangle$ for a specific $w$?

*What space is $w$ from? Is it in the same $V$ as $v$?


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Continuing with my explanation...
From the linear functional, $T_w(v) = \langle v,w\rangle$, we can define another linear functional $T^*_w(v)$ (the ADJOINT) that takes this UNIQUE $w$ and gives us back the corresponding $v$ with which we defined the first linear functional $T_w(v) =\langle v,w\rangle$
This is where my last few questions come up:


*

*What are the domain and codomains of the adjoint linear functional $T_w^*(v)$?

*Why does $\langle T(v),w\rangle = \langle v,T^*(w)\rangle$? Using the definitions I just gave wouldn't $\langle T(v),w\rangle = \langle \langle v,w\rangle, w\rangle$?


phew! that was long, haha. Any helpp would be appreciated! Thanks you guys!
 A: I think that he starting point is to well understand what a linear functional is and what is the dual space, i.e. the space of linear functionals.
We start with a vector space $V$ over a field $K$ nad we say that a function T$:V\rightarrow K$ is a linear functional iff 
$$T(a \vec u +b \vec v)=aT(\vec u)+bT(\vec v) \qquad \forall\; a,b \in K \quad \mbox{and}\quad \forall \;\vec u, \vec v \in V 
$$
Let $V^*$ the set of all such functionals, ve can prove that it is a vector space when equipped with the addition:
$$
(T+S)(\vec v)= T(\vec v)+S(\vec v)
$$
and the scalar multiplication
$$
(aT)(\vec v)= a \left(T(\vec v)\right)
$$
Such vector space $V^*$ is called the algebraic dual of $V$. Note that his elements are not vectors of $V$ but linear functionals.
Now, if $V$ is an Hilbert space of finite dimension $n$, the Riesz Representation Theorem say us that any element of this space $V^*$ can be represented as the inner product of  $\vec v \in V$ with a given vector $\vec t$, i.e.
$$
T(\vec v)= \langle \vec v, \vec t \rangle
$$
This result can be extended also to infinite dimensional spaces, but only for bounded linear functionals. In this case we have to distinguish the algebraic dual $V^*$ from the continuous (or topological) dual $V'$ that is the subspace $V'\subset V^*$ of continuous (bounded) functionals.
Here I avoid this problem limiting the discussion to finite dimensional spaces, where $V'=V^*$.
Here the key fact is that $T$ and $\vec t$ are two different things: $\vec t$ is a vector of $V$, $T$ is a linear functional i.e. an element of $V^*$. 
To highlight this difference ve can call $T$ a 1-form, or a dual vector, and $\vec t$ is a vector that in a well defined sense given by the RRT correspond to the 1-form $T$. 
We can say that $\vec t$ defined by $\langle \vec v, \vec t\rangle \; \forall \vec v \in V$ is the adjoint of the 1-form $T$. But , in my opinion, it is better to reserve this term only for the adjoint of a linear operator and not use it for the functionals to avoid confusion..
Now we can go to your questions.
1) Why $\vec t$  (your $w$) is unique?
Given $T \in V^*$ such that $\langle \vec x,\vec t \rangle =T(\vec x)$ suppose that there is $\vec t'$ s.t.  $\langle \vec x,\vec t' \rangle =T(\vec x)$
Than we have :
$$
\langle \vec x, \vec t-\vec t' \rangle=\langle \vec x, \vec t\rangle -\langle\vec x,\vec t' \rangle= T(\vec x)-T(\vec x) = 0 \quad \forall \vec x \in V
$$
so , in particular, $\langle \vec t-\vec t',\vec t - \vec t'\rangle=0$ and this implies $\vec t= \vec t'$ .
(2) What space is $w$ from?
As you see $\vec t$ ( your $w$) is  a vector $\in V$, but, as just noticed, it is different form $T$.
About the other questions it seems that you have a bit of confusion deriving, I suppose, from the use of the term adjoint.  The '' adjoint'' ( if you want use this term) of a linear functional $T$ is simply the dual vector $\vec t$ of the 1-form $T$. 
A different thing is the adjoint of a linear operator $A:V \rightarrow V$, that is defined as the operator $A^*$ such that:
$$
\langle \vec x,A^*(\vec y)\rangle=\langle A(\vec x),(\vec y)\rangle
$$
but note that you can not write a thing as $\langle T(\vec x),\vec y\rangle$ since $T(\vec x)$ is a scalar!.
A final note: The notation used here is the notation of mathematicians. Physicians use a bra-ket notation in which vectors are indicated as $\mid v\rangle$ ( kets) and linear functionals  as $\langle t\mid$ ( bra) so that the inner product become $\langle t\mid v \rangle$ that, in my opinion ( maybe because I was a physicist), is more expressive and clear.
