Why is this the basis for the state space of two electrons? An electron can be either spin-up $|\!\uparrow\,\rangle$ or spin-down $|\!\downarrow\,\rangle$. 
Moreover, it can be in a superposition of the two states, i.e. $\alpha \,|\!\uparrow\,\rangle + \beta \,|\!\downarrow\,\rangle$, where $\alpha, \beta \in \mathbb C$ and $\alpha\alpha^*=|\alpha|^2$ is the probability of measuring the electron in the spin-up state, while $\beta\beta^* = |\beta|^2$ is the probability of measuring the electron in the spin-down state.
The state space is a two-dimensional, complex vector space with basis vectors $|\!\uparrow\,\rangle$ and $|\!\downarrow\,\rangle$.
Given two electrons, the state space is given by $H_1 \otimes H_2$ where $H_1$ is the state space of the first electron and $H_2$ is the state space of the second electron. This is a four dimensional, complex vector space with basis vectors $|\!\uparrow\,\rangle \otimes |\!\uparrow\,\rangle$, $|\!\downarrow\,\rangle \otimes |\!\uparrow\,\rangle$, $|\!\uparrow\,\rangle\otimes |\!\downarrow\,\rangle$ and $|\!\downarrow\,\rangle \otimes |\!\downarrow\,\rangle$.
These basis vectors are usually written as $|\!\uparrow\uparrow\rangle$, $|\!\downarrow\uparrow\rangle$, $|\!\uparrow\downarrow\rangle$ and $|\!\downarrow\downarrow\rangle$ respectively. Meaning
$$H_1 \otimes H_2 = \{\alpha|\!\uparrow\uparrow\rangle +\beta|\!\downarrow\uparrow\rangle +\gamma |\!\uparrow\downarrow\rangle +\delta|\!\downarrow\downarrow\rangle:\alpha,\beta,\gamma,\delta \in \mathbb C\}$$
Why is the basis usually taken to be
$$|\!\uparrow\uparrow\rangle, \ \ \tfrac{1}{\sqrt 2}\left(|\!\uparrow\downarrow\rangle+|\!\downarrow\uparrow\rangle\right)
, \ \ \tfrac{1}{\sqrt 2}\left(|\!\uparrow\downarrow\rangle-|\!\downarrow\uparrow\rangle\right), \ \ |\!\downarrow\downarrow\rangle
\tag{1}$$
I see that these are orthonormal with repsect to the Hermitian inner product, but aren't
$$|\!\uparrow\uparrow\rangle, \ \ |\!\downarrow\uparrow\rangle, \ \ |\!\uparrow\downarrow\rangle,  \ \ |\!\downarrow\downarrow\rangle
\tag{2}$$
 A: The first, second and last states in (1) are actually a basis for the triplet of $S=1$ states, while the third in (1) is the singlet $S=0$ state.  
As noted by @Paul, the triplet states are also symmetric under permutation of the spins, while the singlet is antisymmetric.
In general, Schur-Weyl duality is a deep mathematical result that allows the construction of states in the $n$-fold tensor product $H_1\otimes H_2\ldots \otimes H_n$ so they also transform by irreducible representations of the permutation group $S_n$.  
In the case of two spin-$1/2$ particle as you have here, the appropriate permutation group is $S_2$.  The two irreducible representations of $S_2$ are the fully symmetric and fully antisymmetric (or alternating) representations.
If you had $3$ spin-$1/2$ particles, you could arrange them in irreps of $S_3$: one fully symmetric irrep containing the $S=3/2$ states, and two irreps of mixed symmetric each containing $S=1/2$ states.
This decomposition does NOT depend on the Hamiltonian.
