Let $(X,\|\cdot\|)$ be a normed linear space and define $\psi_f\in X^{**}$ by the equation $\psi_f(\varphi)=\varphi(f)$ for every $\varphi\in X^*$. Claim: $\|\psi_f\|_{**}=\|f\|$.
Proof: $\|\psi_f\|_{**}\le \|f\|$ is trivial from the definition.
For the reverse I tried thinking of a bounded linear functional $\varphi$ such that $|\psi_f(\varphi)|=|\varphi(f)|=\|f\|$. But for a general space, I can't think of many interesting bounded linear functionals that are always in the dual space. There's the zero functional, but that doesn't help here. I thought about the norm itself, since this is a functional, but it's not necessarily linear.
Is there a linear functional such that $\varphi(f)=\|f\|$ and we somehow linearly extend this to all other points in $X$? Obviously we could define $\varphi(af) = a\|f\|$ for every $a\in\Bbb R$. It would be nice to have an inner product in order to decide which vectors are orthogonal to $f$ and map them all to zero, and then use that to define $\varphi$ on a general vector. But if we only have a norm, I don't know of a way to obtain such an inner-product. I'm only familiar with going the other way, using the inner-product to say what the norm is.
In general, the won't be an inner product inducing the norm (think for example of the $\|\cdot\|_1$-norm on $\mathbf R^2$).
But there is the Hahn-Banach theorem giving you the extension you need. In the form you need it here:
The proof uses the following two steps: Some tedious calculations show that you can always extend for one extra dimension, then a Zorn's lemma argument gives you a maximal extension, what be the first step must be an extension to the whole of $X$.
To your problem: Define $\phi$ on $\mathbf R f$ as you did ($\phi(af) = a \|f\|$) and extend to $X$ using Hahn-Banach.