My answer is in the context of $\mathbb{R}^n$, and I will work directly with the usual definitions of basis, span, and linear independence. I believe doing it this way is helpful because this is the perspective of many students taking linear algebra (especially those in a first course): working in Euclidean space, and wanting to understand how the intuition relates directly to the abstract definitions.
What is a basis of $\mathbb{R}^n$? It is a set of linearly independent vectors that spans $\mathbb{R}^n$. But then (1) what does being linearly independent mean, and (2) what does it mean to span $\mathbb{R}^n$?
To answer (1), let’s consider a set of vectors $\vec{v_1}, \vec{v_2}, …, \vec{v_k}$ in $\mathbb{R}^n$. To say that $\vec{v_1}, \vec{v_2}, …, \vec{v_k}$ are linearly independent means that the only linear combination of these vectors that give the zero vector is the trivial solution. In other words, if you have $a_1\vec{v_1} + v_2\vec{v_2} + … + a_k\vec{v_k} = \vec{0}$ for some real numbers $a_1, a_2, …, a_k$, then we must have $a_1 = a_2 = … = a_k = 0$. So what does this tell us? Well, in particular it tells us that each of the vectors $\vec{v_1}, \vec{v_2}, …, \vec{v_k}$ cannot be written as a linear combination of the others. For example, $\vec{v_1}$ cannot be written as $b_2\vec{v_2} + ... + b_k\vec{v_k}$ no matter which real numbers $b_2, ..., b_k$ you choose. Why? Because, if we could, then we will have $\vec{v_1} = b_2\vec{v_2} + ... + b_k\vec{v_k}$, and so subtracting the right-hand side from both sides, we get $\vec{v_1} - b_2\vec{v_2} - ... - b_k\vec{v_k} = \vec{0}$. Note that the coefficient of $\vec{v_1}$ is 1, which is not 0. So we have a non-trivial solution to the equation $a_1\vec{v_1} + v_2\vec{v_2} + … + a_k\vec{v_k} = \vec{0}$, contradicting the assumption of linear independence.
So, intuitively, in $\mathbb{R}^2$, if two vectors $\vec{u}, \vec{v}$ are linearly independent, this means $\vec{u}$ cannot be written as $c\vec{v}$ no matter what real number $c$ you choose (and vice versa, $\vec{v}$ cannot be written as $c\vec{u}$). This means $\vec{u}$ doesn't lie in the line that is formed by $\vec{v}$. In $\mathbb{R}^3$, if three vectors $\vec{u}, \vec{v}, \vec{w}$ are linearly independent, this means any one of these vectors cannot be written as a linear combination of the others. For example, $\vec{u}$ cannot be written as $c\vec{v} + d\vec{w}$, in other words, $\vec{u}$ doesn't lie in the plane spanned by $\vec{v}$ and $\vec{w}$. (You can see my answer here for some geometric intuition as to why two linearly independent vectors in $\mathbb{R}^3$ span a plane.) So if you have three vectors lying in the same plane, you know they cannot be linearly independent.
To answer (2): if the span of $\vec{v_1}, \vec{v_2} ..., \vec{v_k}$ equals $\mathbb{R}^n$, what does this mean? It means that the set of all linear combinations $a_1\vec{v_1} + a_2\vec{v_1} + ... + a_k\vec{v_k}$ equals $\mathbb{R}^n$. You can think of spanning $\mathbb{R}^n$ as sort of like a game. If I give you any vector $\vec{v}$ in $\mathbb{R}^n$, can you give me scalars $a_1, a_2, ..., a_k$ so that $\vec{v} = a_1\vec{v_1} + a_2\vec{v_2} + ... + a_k\vec{v_k}$? In other words, if you give me any vector $\vec{v}$ in $\mathbb{R}^n$, can you write it as a linear combination of $\vec{v_1}, \vec{v_2}, ..., \vec{v_k}$? Since $\vec{v_1}, \vec{v_2}, ..., \vec{v_k}$ spans $\mathbb{R}^n$, the answer is yes.
So, what then does it mean if $\vec{v_1}, \vec{v_2}, ..., \vec{v_k}$ is a basis for $\mathbb{R}^n$? It means none of these vectors can be written in terms of the others (i.e. linear independence), and it means that by using linear combinations of these vectors, I can reach any vector in $\mathbb{R}^n$ (i.e. spans $\mathbb{R}^n$).
Linear independence sort of tells you that you don't have any ''extra/unnecessary'' vectors. To be linearly dependent (i.e. not linearly independent) means some vector is in the span of the other vectors, so it is ''extra/unnecessary'' in the sense that we don't need it. Whenever we encounter this vector, we can just replace it as a linear combination of the other vectors. So really all we need are those other vectors.
So, a basis of $\mathbb{R}^n$ is a set of vectors that is able to reach every vector in $\mathbb{R}^n$, but at the same time we don't have any ''extra/unnecessary'' vectors. For example, if we have three vectors in $\mathbb{R}^2$ no two of which lie in the same line, then any two of these vectors will span a plane, i.e. all of $\mathbb{R}^2$. So the third vector is extra/unnecessary. In other words, these three vectors would span $\mathbb{R}^2$, but it is not linearly independent. So they do not form a basis for $\mathbb{R}^2$. If we have two linearly independent vectors in $\mathbb{R}^3$, then they span a plane, which is not all of $\mathbb{R}^3$. So they would be linearly independent, but not span $\mathbb{R}^3$. So they do not form a basis for $\mathbb{R}^3$. A basis is when both these conditions are met. For example, two linearly independent vectors in $\mathbb{R}^2$ form a basis for $\mathbb{R}^2$, and three linearly independent vectors in $\mathbb{R}^3$ form a basis for $\mathbb{R}^3$. More generally, $n$ linearly independent vectors in $\mathbb{R}^n$ form a basis for $\mathbb{R}^n$.