I am assuming that you are working in some Hilbert space $\mathbb{H}$. The key fact is that any continuous linear functional $f:\mathbb{H} \to \mathbb{C}$ can be represented by a unique element $\phi \in \mathbb{H}$ in the sense that $f(x) = \langle \phi, x \rangle$ for all $x \in \mathbb{H}$ (and conversely any element $\phi \in \mathbb{H}$ determines a unique continuous linear functional on $\mathbb{H}$). A little work shows that $\|f\| = \|\phi\|$. The big deal above is the uniqueness.
Now suppose $T: \mathbb{H} \to \mathbb{H}$ is a continuous linear operator and $y \in \mathbb{H}$. Then $f_y(x) = \langle y, Tx \rangle$ is a continuous linear functional, and can be represented by some $\phi_{y} \in \mathbb{H}$, ie, $f_y(x) = \langle \phi_{y}, x \rangle$. It is easy to show using uniqueness that $\phi_{\lambda y} = \lambda \phi_y$ and $\phi_{y_1+y_2} = \phi_{y_1}+\phi_{y_2}$, hence the mapping $y \mapsto \phi_y$ is linear. Furthermore, since $|f_y(x)| = |\langle \phi_{y}, x \rangle| \leq \|x\| \|y\| \|T\|$, choosing $x=\frac{\phi_y}{\|\phi_y\|}$ (or zero, if $\phi_y=0$) shows that $\| \phi_y \| \leq \|T \| \|y\|$, hence the mapping $y \mapsto \phi_y$ is bounded, and hence continuous. Instead of writing $y \mapsto \phi_y$, we now use the more usual notation $T^* y = \phi_y$. We have shown that $T^*$ is linear, continuous, and completely defined by the requirement that $\langle T^*y, x \rangle = \langle y, Tx \rangle$ for all $x,y \in \mathbb{H}$.
All other properties follow from this requirement and properties of the inner product.
For example, to show additivity: $\langle (S+T)^*y, x \rangle = \langle y, (S+T)x \rangle = \langle y, Sx \rangle + \langle y, Tx \rangle = \langle S^*y, x \rangle + \langle T^*y, x \rangle = \langle (S^*+T^*)y, x \rangle$, hence $(S+T)^* = S^*+T^*$.
For the adjoint of the adjoint: $\langle (T^*)^*y, x \rangle = \langle y, T^*x \rangle = \overline{\langle T^*x,y \rangle} = \overline{\langle x,Ty \rangle} = \langle Ty,x \rangle$, hence $(T^*)^* = T$.
Now try the rest.