The assertion "Every vector space has a basis" implies the axiom of choice in ZF. This means that without the axiom of choice there are spaces that have no basis.
It is open at the moment whether or not this implies that there is a space with a trivial dual as well. However, it is consistent that there is a space whose dual is trivial. That is to say that the only linear functional is indeed $0$.
The usual construction yields a vector space $W$ with the following properties:
- $W$ is not spanned by any finite set (i.e. it does not have a finite dimension),
- Every proper subspace has a finite dimension,
- The only linear transformation from $W$ into itself are multiplication by scalars from the field. In particular this implies that there is only one linear functional: the zero functional.
Let us consider this case: Let $V$ be $\mathbb R^n$ for some $n$, and let $W$ be as described above (with the field being $\mathbb R$).
Choose $n$ vectors, $w_1,\ldots,w_n$, which are linearly independent (this process requires no axiom of choice since it can be described in finitely many steps). Let $W'$ be the span of $w_1,\ldots,w_n$.
There is a natural map $f$ from $V$ into $W'$, and hence into $W$. Consider $f^*$, its domain is $W^*=\{0\}$. It is far far from being onto $V$.
Wait, it gets worse. Consider $f\colon V\to M=W\oplus \mathbb R^{n-1}$ as the map which maps $e_1\to w\in W$ (which is nonzero) and $e_2,\ldots,e_n$ into the $\mathbb R^{n-1}$ part.
Every functional on $M$ is $0$ on $W$ and a usual functional on $\mathbb R^{n-1}$. However there is no surjective map from $M^*$ onto $V^*$, simply since $\dim(M^*)=n-1<\dim(V^*)=n$.
Any linear map $fu$ from $V=\mathbf R^3$ to $\mathbf R$ is determined by its values on the vectors of a base $\mathcal B =(e_1, e_2, e_3)$. For if $v=\lambda e_1+\mu e_2+\nu e_3$, then $f(v)=\lambda f(e_1)+\mu f(e_2)+\nu f(e_3)$.
Now if $f(e_1)=\alpha_1$, $f(e_2)=\alpha_2$, $f(e_3)=\alpha_3$ and if $e_1^*, e_2^*,e_3^*$ is the dual basis of $\mathcal B$, it's easy to check that
$$f=\alpha_1 e_1^*+\alpha_2 e_2^*+\alpha_3 e_3^*$$ since both sides take the same value for $\,e_1,e_2,e_3$.
Best Answer
More generally, suppose we have a linear transformation $T : V \to W$. Now let $f \in W^*$, the dual of $W$. That is, we have
$$V \xrightarrow{T} W \xrightarrow{f} \mathbb{F}$$
where $\mathbb{F}$ is the base field of $V$ and $W$. You can 'pull back' $f$ to be defined on $V$, rather than $W$, but still mapping into $\mathbb{F}$. This is acheived by precomposing with $T$ which gives an element of $V^*$. That is why we call the map $W^* \to V^*$, $f \mapsto f\circ T$ the pullback.
Heuristically, because we have a linear transformation $T$ which maps from $V$ to $W$, you can think of $V$ as the starting point, and $W$ as the destination. If we then have some object defined on $W$ (like an element of the dual space), we can try and pull it back from the destination to the starting point by somehow using $T$.
If I knew how to draw commutative diagrams, I'd include one here.