I know basically the difference between Klein-Gordon and Dirac field is spin. But I am not sure where we need to implement this info.
The solutions of both equations are the wave packets which includes the sum of creation and annihilation operators.
$$ \Psi(x) = \int [a_pe^{-ipx} + a_p^\dagger e^{ipx}]\,d^3 p. $$
where we use the spin information? Why should I use Klein Gordon for spin-0 and Dirac equation for 1/2?
The momentum decomposition you wrote is valid only for a scalar (spinless), real field, satisfying the Klein-Gordon equation.
When considering a field with spin, like a spin-$1/2$ field satisfying the Dirac equation, you must include the polarization vectors, obtaining something of the form $$ \psi_\alpha(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}}[ c_s(\textbf{p}) u_\alpha(s,\textbf{p}) e^{-ipx} + d_s^\dagger(\textbf{p}) v_\alpha(s,\textbf{p}) e^{ipx} ] $$ $$ \bar \psi_\alpha(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}}[ c_s^\dagger(\textbf{p}) \bar u_\alpha(s,\textbf{p}) e^{ipx} + d_s(\textbf{p}) \bar v_\alpha(s,\textbf{p}) e^{-ipx} ] $$ where $c,c^\dagger$ and $d,d^\dagger$ are respectively annihilation/creation operators of particles and corresponding antiparticles, $u,v$ are the polarization vectors, which encode the information about the spin, and $N_{s,\textbf{p}}$ normalization factors depending on the convention used. The sum is extended over all momentum ($\textbf{p}$) and spin ($s$) eigenstates. The objects I denoted with $ue^{-ipx}$ and $ve^{ipx}$ are called Dirac spinors. They have four components, reflecting the fact that a Dirac field describes both electron and positron, each having 2 spin degrees of freedom (for a total of $2+2=4$ degrees of freedom).
In other words, the spin information is encoded in the additional degrees of freedom of the field, in this case the spin-index which I denoted with $\alpha$. This additional degrees of freedom do not evolve independently, as you can see from the (free) Dirac equation, which showing explicitly the spinor indices reads $$ ( i \gamma^\mu_{\alpha\beta} \partial_\mu - m \delta_{\alpha\beta})\psi_\beta(x) = 0, \quad \forall \alpha=1,2,3,4,$$ where $\gamma^\mu$ are the gamma matrices, and a sum over the spin-index $\beta$ is implicit.
For another example you can look at spin-1 fields (e.g. photons). In this case the quantum field is denoted by $A_\mu(x)$ with $\mu$ a vector index, which is the spin-index for spin-1 fields. The decomposition has now the form: $$ A_\mu(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}} [ a(s,\textbf{p}) \varepsilon_\mu(s,\textbf{p}) e^{-ipx} + a^\dagger(s,\textbf{p}) \varepsilon_\mu^*(s,\textbf{p}) e^{ipx} ]. $$ Again, $s$ denotes the spin states (which are usually called polarization states in this context), and $\varepsilon$ are the polarization vectors.
Finally, note that each spin component of the Dirac field satisfied the Klein-Gordon equation: $$ (\square + m^2)\psi_\alpha(x) = 0, \forall \alpha $$ An excellent explanation of this can be found here.
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