The statement that $a|0\rangle=0$ is exactly the same as the statement $\langle 0|a^\dagger=0$. This is one of these things that is easier to see in the mathematicians notation. In their notation, the inner product between two vectors $u$ and $v$ is denoted by $\langle u,v\rangle$. The adjoint of an operator $A$ is defined by $\langle u, Av\rangle=\langle A^\dagger u,v\rangle$. Now, let us denote the vector $|0\rangle$ by $u$. Then for every vector $v$, which in ket-bra notation we may denote by $|v\rangle$, we have $$\langle 0|a^\dagger|v\rangle=\langle u, a^\dagger v\rangle=\langle au,v\rangle=\langle 0,v\rangle=0.$$
The first equality is just a change of notation. The second is the definition of adjoint of an operator. The third is the fact that the annihilation operator annihilates the vacuum, i.e. $a|0\rangle=0$, or, in our notation, that $au=0$. Here a word of caution: when going between the mathematicians and the physicists language one must be careful not to confuse the vacuum $|0\rangle$ with the $0$ of the Hilbert space. This is why we had to introduce the new label $u$. Finally, the fourth equality is a standard property of inner products. It can be easily derived from the skew linearity $\langle k u,v\rangle=\overline{k}\langle u,v\rangle$ for all vectors $u$ and $v$ and scalars $k$. In particular, take $k=0$.
We conclude that for all $|v\rangle$ we have $\langle 0|a^\dagger|v\rangle$. Therefore, the operator $\langle 0|a^\dagger=0$. Indeed, an operator is equal to $0$ if and only if its action on all of the vectors vanishes.