Let me answer your last two questions first:
In the example of $Q[t]$, what I would call the "standard basis" is $\{1, t, t^2, t^3, \dots \}$. This is a basis because every polynomial in $Q[t]$ can be uniquely expressed as a linear combination of those elements. But there are many other (in fact, infinitely many) bases for $Q[t]$; for example, $\{1, 1+ t, 1 + t +t^2, 1 + t + t^2 + t^3, \dots \}$ is also a basis because, again, every polynomial can be uniquely expressed as a linear combination of those elements. (As an exercise, can you express, say, $t^2 +3t -1$ as a linear combination of the elements in that basis?) Remember, vector spaces generally have many different bases.
Now on to your first question. As other answers have said, every vector space $V$ has a basis, which is a set of vectors $B$ such that every vector in $V$ can be uniquely written as a linear combination of vectors in $B$. (Note: by definition, linear combination means a finite sum.)
Unfortunately, for a lot of commonly occurring infinite dimensional spaces (e.g., function spaces such as the one you mentioned), there's no good way of writing down an explicit basis.
One last note: when one studies Hilbert spaces (which are "complete inner product spaces"), one uses the term "basis" (or "orthonormal basis") to mean something subtly different from our use of "basis" above. In that context, you are allowed to take infinite sums of "basis" elements. This confused me when I first studied Hilbert spaces.
First, not all functions from the reals to the reals are "polynomial, triginometric, exponential". These form in fact a very tiny subset of all functions. A function is not a formula. It's just that for every $x \in \mathbb{R}$ there exists a unique value $f(x) \in \mathbb{R}$. The unicity is what makes it a function. The asignment of $f(x)$ to $x$ is completely arbitrary, e.g. I could take the 7-th digit in the decimal (infinite) representation of $x$ for all $x > \sqrt{17}$, the 5-th digit of that representation for all $x < \sqrt{17}$ and $f(\sqrt{17}) = \pi$. And this is even relatively nice, because I can write a program for it, but this need not be the case in general. A truly "random" function would just be an infinite list of values, one for each real, assigned all independently of each other. So there is no hope for Taylor expansions etc. Realise that $\mathbb{R}^\mathbb{R}$ is truly a huge set.
The operations on $\mathbb{R}^\mathbb{R}$ are pointwise, so if we have two such functions $f,g$ then we define $f+g$ as a new function, by telling what its value on an arbitrary $x \in \mathbb{R}$ is: $(f+g)(x) = f(x) + g(x)$, i.e. we just add the values of $f$ and $g$ at $x$. Similarly for a scalar $c \in \mathbb{R}$, we define $(c\cdot f)(x) = cf(x)$ for all $x \in \mathbb{R}$, where the latter is just standard multiplication in $\mathbb{R}$. The $0$ is just the function where all values are equal to $0$.
Both have infinite sets of linearly independent elements (or vectors, as elements of a vector space are called, even though they are not "vectors" in the old fashioned sense, like the functions in $\mathbb{R}^\mathbb{R}$): take the functions $f_p$, defined for a fixed $p \in \mathbb{R}$: $f_p(x) = 1$ if $x = p$, and $f_p(x) = 0$ if $x \neq p$. So all $0$ except for a spike at $p$.
Why are the $f_p$ linearly independent? By definition, we need to consider a finite linear combination of distinct $f_{p_i}$: $c_{p_1} \cdot f_{p_1} + \ldots + c_{p_n} \cdot f_{p_n} = 0$ (equality as functions, which means just that they have the same values for all $x$), and we need to show that then all $c_{p_i}$ are $0 \in \mathbb{R}$. Because the equality holds for all $x$, we can use $x = p_1$ in particular. Then $$(c_{p_1} \cdot f_{p_1} + \ldots + c_{p_n} \cdot f_{p_n})(p_1) = c_{p_1}f_{p_1}(p_1) + \ldots + c_{p_n} f_{p_n}(p_1) = 0$$ As $f_{p_2}(p_1) = 0$, as $p_2 \neq p_1$, and so on, but $f_{p_1}(p_1) = 1$ this comes down to $c_{p_1} = 0$. The same idea works for all other coefficients as well. So the set of $f_p$ is linearly independent.
The same idea works in $\mathbb{R}^\mathbb{N}$, the set of sequences, which look more like normal vectors, but of infinite length. We just see this as the set of functions from $\mathbb{N}$ to $\mathbb{R}$ and the same operations and independent functions apply. So both spaces are infinite-dimensional.
If we see $\mathbb{R}^\mathbb{N}$ as a set of functions (as we should) then the function $T$ is just the restriction of $f$ to $\mathbb{N}$. To see that this is linear, take $f,g \in \mathbb{R}^\mathbb{R}$. Then $T(f+g)$ is defined for all $n$ as $(f+g)(n)$, which is in turn defined as $f(n) + g(n)$, and this equals $T(f)(n) + T(g)(n)$, as $T$ does "nothing", it's just the restriction of a function to a smaller domain. The latter sum is just by definition $(T(f) + T(g))(n)$, and as this holds for all $n$ we have equality of functions (or sequences, because we index by $\mathbb{N}$) and $T(f+g) = T(f) + T(g)$. The same can be done for scalars as well.
A matrix for a linear map $T$ is formed by choosing bases for both spaces and computing the base expansion for every $T(b)$ for all basis elements in the domain (as the columns). But here a base for $\mathbb{R}^\mathbb{R}$ is uncountable, so we cannot write it down as a matrix: these can be at most countably infinite in both dimensions.
Best Answer
First of all, it is not the case that the Taylor series of a function necessarily converges to the function, even if all derivatives at the point are defined. One classic example of this is the function $$ f(x) = \begin{cases} e^{-1/x^2} & x \neq 0\\ 0 & x=0 \end{cases}. $$ Though it might not be obvious that this is the case, it turns out that this $f(x)$ has infinitely many derivatives defined at $x = 0$. In fact, we have $f^{(k)}(0) = 0$ for every $k$. Thus, the $k$th Taylor polynomial centered at $0$ is given by $T_k(x) = 0$, and is clear that the series of functions $T_k$ fails to converge to $f$ in any reasonable sense.
You might find it interesting that the function $g(x) = e^{-x^2}f(x)$ exhibits similar behavior at $x = 0$ and is also square-integrable over $\Bbb R$.
A weaker version of your statement does hold: if we consider $L_2([a,b])$, the set of square integrable functions $f:[a,b] \to \Bbb C$, then for any $f \in L_2$ there exists a sequence of polynomials $p_n(x)$ such that $p_n \to f$ relative to the $L_2$ norm. That is, there exists a sequence of polynomials $p_n$ such that the integral $\int_a^b |p_n(x) - f(x)|^2 dx$ converges to zero as $n \to \infty$. This can be proven as a consequence of the Weierstrass approximation theorem. In other words, we might say that the subspace of polynomials in $L_2([a,b])$ is dense.
Note that $L_2(\Bbb R)$, the set of square integrable functions $f:\Bbb R \to \Bbb C$, fails to contain any non-zero functions of the form $a_0 + a_1 x + \cdots + a_n x^n$, so I would argue that it no longer makes sense to approximate $f(x)$ with polynomials in this context. That being said, results regarding the approximation of functions $f \in L_2(\Bbb R)$ exists and are of vital importance in harmonic analysis.