What does $\mathbb{R}^d$ mean, how does one denote different basis vectors for it and projections to different vector spaces? This might sound like a very basic question, but what does a vector mean? What does $\mathbb{R}^d$ mean and how do you denote which basis you are referring too? Do you always assume the "usual" basis $e_1 = (1, 0, ... , 0)$, $e_i$, etc? 
For example consider:
$$x =  
\left(
\begin{array}{c}
x_1\\
\vdots \\
x_i\\
\vdots \\
x_d\\
\end{array}
\right) 
$$
where $x_i \in \mathbb{R}$. The way I interpreted that was that $x$ lied in some space (with d basis vectors) and each index of the vector indicated how much you need to move according to the basis vector that corresponds to that "coordinate". However, I was a little unsure about something. How do I know which basis I am working on? Because the vector x could bet two completely different things depending on what is the basis.
Do we always have to assume we are working with the "usual" basis? i.e. $e_1 = (1, 0, ... , 0)$ etc?
The issue that I was concerned was that I wanted to define a linear transformation that mapped elements in some space $X$ to elements in any new space $Y$ (with maybe a "weird" basis, however, of some fixed size).
My initial idea was to define the following linear transformation:
$$T(x) = 
\left(
\begin{array}{c}
\langle x, a_1 \rangle\\
\vdots \\
\langle x, a_k \rangle\\
\vdots \\
\langle x, a_d \rangle\\
\end{array}
\right)=
\left(
\begin{array}{c}
x^Ta_1\\
\vdots \\
x^Ta_i\\
\vdots \\
x^Ta_d\\
\end{array}
\right) 
$$
which seemed like a good idea to me at first because each component indicates the projection in the direction of $a_i$ (assuming they are unit vectors). However, I realized that it only worked if people knew before hand which basis I was referring to and that was what I meant by a "coordinate". Hence I was curious to see what was the precise way to do something like this. Also, do we say that such a transform maps things in the original space X to $\mathbb{R}^d$? Or how do you rigorously express how the map works? Does it span $\mathbb{R}^d$?
I know that a map such as:
$$ M(x) = \sum^{d}_{i=1} (x^Ta_i)a_i$$
would technically work, but seemed a little ugly.
Is there a way to define a linear transformation that maps elements to a new basis vector? I really wanted to use something similar to what defined as $T(x)$ but as unsure if that was legal or how to make it precise/rigorous.
The reason I really liked my T(x) was because it seemed like a very elegant, compact way to encode that information and also had the potential to be expressed as a product of matrices. For example, consider a set of vectors of interest $x_i$ and let them be the rows of the matrix X. Then I think the following equation expresses T(x) using matrix multiplication:
$$ T = XA $$
where the columns of A are the $a_i$'s. This seems like a really nice way of expressing T(x).

I have taken linear algebra before but it was nearly 4 years ago. Sorry if this is something I was suppose to remember.
 A: 
what does a vector mean?

A vector is an element of a vector space. I think your confusion comes from the fact that every two vector spaces with the same (finite) dimension are isomorphic. In particular, every $n$ dimensional $K$-vector space is isomorphic to $K^n$.

How do I know which basis I am working on? Because the vector x could bet two completely different things depending on what is the basis.
Do we always have to assume we are working with the "usual" basis? i.e. $e_1=(1,0,...,0)$ etc?

Unless it's said explicitly, you're using the canonical basis. If that's not the case, the common notation is that given a basis, $\beta$, $[x]_\beta$ is the vector $x$ written in that basis.

The issue that I was concerned was that I wanted to define a linear transformation that mapped elements in some space X to elements in any new space Y (with maybe a "weird" basis, however, of some fixed size).

Consider the bases, $\alpha = \{a_1,a_2,\dots,a_n\}$, of $X$ and, $\beta=\{b_1,\dots,b_m\}$, of $Y$. In order to define a linear transformation from $X$ to $Y$, it's enough to determine what does the transformation do to $\alpha$. Once you know this, you can extend it by linearity, try proving this.
Define a linear transformation $T:X\to Y$ as
$$
T(a_i) = \sum_{j=1}^m x_{ij}b_j,
$$
where $x_{ij}\in K$. The matrix $\mathbb{T}_\alpha^\beta = (x_{ij})^T$ is the associated matrix of the transformation. 
Observe that the coefficients $x_{ij}$ depend completely in the choice of bases. To be able to use this matrix, we must express the vectors of $X$ in terms of the basis $\alpha$,  and the vectors of $Y$ in terms of $\beta$, in which case $[a_i]_\alpha = (0,\dots, 0,1,0,\dots 0)$ and
$$
\mathbb{T}_\alpha^\beta [a_i]_{\alpha} = \sum_{j=1}^mx_{ij}[b_j]_\beta.
$$
I think most of your doubts would be clarified by reading the first chapters of your favorite linear algebra book. 
