You're conflating transcendence basis and vector space (Hamel) basis.
If $K$ is an extension field of $F$, there exists a set $T$ consisting of transcendental elements of $K$ such that
- $T$ is algebraically independent;
- $K$ is algebraic over $F(T)$.
Such a set is a transcendence basis of $K$ over $F$.
A set $U\subseteq K$ is algebraically independent if and only if, for every finite subset $S=\{a_1,a_2,\dots,a_n\}$ (elements pairwise distinct) of $U$, there is no nonzero polynomial $f(X_1,\dots,X_n)\in F[X_1,\dots,X_n]$ such that $f(a_1,\dots,a_n)=0$.
Using a technique very similar to the proof of existence of vector space basis, one can show that a transcendence basis exists and that any maximal algebraically independent set is such a basis; also, two transcendence bases have the same cardinality.
So let $T$ be a transcendence basis of $K$ over $F$. It is not possible to prove that $K=F(T)$, when $T$ is a transcendence basis. To see why, consider the simple case when $K=F(t)$, where $t$ is transcendental over $F$; then also $\{t^2\}$ is a transcendence basis, but clearly $K\ne F(t^2)$. On the other hand, $t$ is algebraic over $F(t^2)$, so $K$ is indeed algebraic over $F(t^2)$. Perhaps more simply, if $K$ is algebraic over $F$, then the empty set is a transcendence basis, but $K$ need not equal $F$.
In general, if $T$ is nonempty transcendence basis and you replace one of its elements, say $t$, by its square $t^2$, then the set $T'$ so obtained is still a transcendence basis, but definitely $F(T')\subsetneq F(T)$. So condition 2 above cannot be replaced by $K=F(T)$.
In the finite case $K=F(t)$ it is certainly false that a transcendence basis is also a vector space basis: indeed, $K=F(t)$ is infinite dimensional as vector space over $F$.
In the case of $F=\mathbb{Q}$ and $K=\mathbb{R}$ a transcendence basis must have the same cardinality as $\mathbb{R}$, by a cardinality argument: we have
$$
\mathbb{Q}(T)=\bigcup_{\substack{S\subseteq T\\S\text{ finite}}}\mathbb{Q}(S)
$$
and every subfield $\mathbb{Q}(S)$ is countable.
Also a vector space basis has the same cardinality as $\mathbb{R}$, but a transcendence basis cannot be a vector space basis. Let $T'=T\setminus\{t\}$, where $t\in T$ is fixed. Then $\mathbb{Q}(T)=F(t)$, where $F=\mathbb{Q}(T')$. Then $\mathbb{Q}(T)$ is infinite dimensional over $F$. If you remove one element from a vector space basis, the subspace spanned by the new set has codimension one.
What you can say is that every $r\in\mathbb{R}$ is algebraic over $\mathbb{Q}(S)$, where $S$ is some finite subset of $T$. Indeed, if $f$ is the minimal polynomial of $r$ over $\mathbb{Q}(T)$, it has finitely many coefficients, so it belongs to $\mathbb{Q}(S)[X]$, for some finite $S\subseteq T$.
For completeness, $\mathbb{R}$ is not a purely transcendental extension of $\mathbb{Q}$, that is, for every transcendence basis $T$ of $\mathbb{R}$ over $\mathbb{Q}$, $\mathbb{Q}(T)\subsetneq\mathbb{R}$. Indeed if equality holds, any permutation of $T$ would induce an automorphism of $\mathbb{R}$, but $\mathbb{R}$ has only the identity automorphism.
Best Answer
An extension $L/K$ is purely transcendental if there exists some algebraically independent set $S$ such that $L = K(S)$.
Every extension $L/K$ can be written as $K(S)$ for some set $S$ (e.g. $S = L$) but maybe or maybe not with $S$ being algebraically independent.
For instance, with $\mathbf{Q}(x) / \mathbf{Q}$ I can take $S = \{x, x^2\}$ or $S_2 = \{x\}$. One of these is algebraically independent and the other is not.
Or $\mathbf{Q}(\sqrt{2}, x)/\mathbf{Q}$ I can take $S_1 = \{\sqrt{2}, x\}$ or $S_2 = \{\sqrt{2}, x + \frac37\sqrt{2}, x^2, \frac{1 - x + x^2}{1 - \sqrt{2}}\}$ but never with $S$ being algebraically independent.