The expression you wrote for $\sigma_z^1 + \sigma_z^2$ is not quite right, but it's not surprising that you're unsure of how to proceed because the notation is somewhat obscuring the real math behind all of this. What's actually going on here is manipulations with tensor products of Hilbert spaces.
The spin state of a single spin-$\frac{1}{2}$ particle is an element of a two-dimensional Hilbert space, let's call it $\mathcal H_\frac{1}{2}$. The spin state of a system consisting of two spin-$\frac{1}{2}$ particles is $\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}$ called the tensor product of $\mathcal H_\frac{1}{2}$ with itself.
If $|+\rangle$ and $|-\rangle$ are the usual spin up and spin down basis elements for $\mathcal H_\frac{1}{2}$, namely eigenvectors of $\sigma_z$ with eigenvalues $\pm\frac{\hbar}{2}$
$$
\sigma_z|+\rangle = \frac{\hbar}{2}|+\rangle, \qquad \sigma_z|-\rangle = -\frac{\hbar}{2}|-\rangle
$$
then a basis for the tensor product $\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}$ consists of all tensor products of basis elements of $\mathcal H_\frac{1}{2}$ of which there are $4$ in this case;
$$
|+\rangle\otimes|+\rangle, \qquad |+\rangle \otimes|-\rangle, \qquad |-\rangle \otimes |+\rangle, \qquad |-\rangle \otimes|-\rangle
$$
In particular, note that the dimension of the vector space of the two-particle spin $\frac{1}{2}$ system is twice the dimension of the single particle spin system.
This far, we have set the stage for defining what $\sigma^1_z$ and $\sigma^2_z$ are. The superscripts simply mean that the operator in question only acts on either the first or the second factor in a tensor product state depending on whether the superscript is a $1$ or a $2$. For example
$$
\sigma_z^1(|+\rangle\otimes|-\rangle) = (\sigma_z |+\rangle)\otimes|-\rangle = \frac{\hbar}{2}|+\rangle\otimes|-\rangle
$$
while
$$
\sigma_z^2(|+\rangle\otimes|-\rangle) = |+\rangle\otimes(\sigma_z|-\rangle) = -\frac{\hbar}{2}|+\rangle\otimes|-\rangle
$$
and in fact one writes
$$
\sigma^1_z = \sigma_z\otimes I, \qquad \sigma_z^2 = I\otimes \sigma_z
$$
which indicates that for example $\sigma_z^1$ acts like $\sigma_z$ in the first factor of the tensor product and like the identity matrix in the second factor, and vice versa for $\sigma_z^2$.
Similarly, if we write down
$$
\vec\sigma^1\cdot\vec\sigma^2 = \sigma_x^1\sigma_x^2 + \sigma_y^1\sigma_y^2+\sigma_z^1\sigma_z^2
$$
then every operator with a $1$ superscript only acts non-trivially on the first factor in a tensor product state while every operator with a superscript $2$ only acts non-trivially on the second factor in a tensor product state.
Best Answer
You missed the vector arrow on $\vec{\alpha}$. So $\vec{\alpha}$ is a "vector" having 3 components ($\alpha_x,\alpha_y,\alpha_z$}. And each of these 3 is a $4\times 4$ matrix.
Or more explicitly: $$\alpha_x=\begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}$$ $$\alpha_y=\begin{pmatrix} 0 & \sigma_y \\ \sigma_y & 0 \end{pmatrix}$$ $$\alpha_z=\begin{pmatrix} 0 & \sigma_z \\ \sigma_z & 0 \end{pmatrix}$$
The Dirac Hamiltonian $$H=c\vec{\alpha}\cdot\vec{p}+\beta mc^2$$ is short-hand notation for $$H=c(\alpha_x p_x+\alpha_y p_y+\alpha_z p_z)+\beta mc^2$$
So finally the Hamiltonian is a $4\times 4$ matrix where each of its components is a differential operator. You can apply this thing to a $4$-component spinor field \begin{pmatrix} \psi_1(\vec{r},t) \\ \psi_2(\vec{r},t) \\ \psi_3(\vec{r},t) \\ \psi_4(\vec{r},t) \end{pmatrix} to obtain another $4$-component spinor field.