Definition: $E$ is called $H^0$-semistable if for all line bundles $L\subset E$ we have $$ h^0(L) \leq \frac{h^0(E)}{2} $$ $E$ is called slope-semistable if for all subbundles we have $$ \frac{\deg L}{\operatorname{rk} L} \leq \frac{\deg E}{\operatorname{rk} E}, $$ i.e. here, because $E$ has rank two $$ \deg L \leq \frac{\deg E}{2}.
Why is $E$, a vector bundle of rank $2$ over a curve $C$, $H^0$-semistable if and only if $E$ is slope-semistable?
The direction slope-semistable implies $H^0$-semistable seems to work by manipulating Riemann-Roch well enough, but I cannot get the other direction to work.
This is not true. By your definition any rank 2 vector bundle $E$ with $H^0(E) = 0$ is $H^0$-semistable; in particular a sufficiently negative twist of any rank 2 vector bundle is $H^0$-semistable. However, slope stability is invariant under any twists, so if you take you favorite slope-unstable bundle and take its sufficiently negative twist, this will be a counterexample.