In many ways $H^1$ is just a natural substitute for $L^1$.
A typical $H^1$ function is a $1$-atom, i.e. a function $\phi\in L^1(\mathbb R^n)$ such that the support of $\phi$ is contained in some ball $B(a,r)$, the bound $$\sup|\phi(\cdot)|\leq \frac{1}{m(B(a,r))}$$ holds true, and $$\int_{\mathbb R^n} \phi(x)dx=0.$$
In fact, following a suggestion by Fefferman, one can use $1$-atoms to characterize the whole Hardy space $H^1$. More precisely, the following is true.
Theorem. A function $f$ belongs to $H^1(\mathbb R^n)$ if and only if there exist $1$-atoms $\phi_k$ and complex numbers $c_k$ such that $$f=\sum_{k=1}^{\infty}c_k\phi_k,$$ where the convergence is in $L^1$ and $\sum_{k}|c_k|<\infty.$
One should note that the mean-value condition in the definition of $1$-atom plays a key role here. If it is removed from the definition, the above description with sums would result in the standard Lebesgue space $L^1$.
$H^1$ is a good testing field to study various classes of multiplier transformations. For instance, the singular integral operators of the form $$Tf=\lim_{\epsilon\to0}\int_{|y|>\epsilon}\frac{K(x-y)}{|x-y|^n}f(y)dy$$ where $K$ is a smooth homogeneous function of degree $0$ such that $\int_{|y|=1}K(y)dy=0$, extend to bounded operators on $H^1$. In particular, the Riesz transforms are bounded on $H^1$. Moreover, an integrable function $f$ belongs to $H^1$ if and only if its Riesz transforms $R_jf$ are also in $L^1$.