computer algebra - Basis for modular forms of half-integral weight

Edit: Here's a rather silly method that should work if SAGE is just giving you cusp forms: $Gamma_0(4)$ has a single normalized cusp form of weight 6, given by $eta(2tau)^{12} = q - 12q^3 + 54q^5 - dots$, so take your basis of cusp forms of weight $k/2 + 6$, and divide each element by this form to get a basis of the space of modular forms of weight $k/2$.

Edit in response to Buzzard: Thanks for pointing out that I should make this argument. Here is a proof that the cusp form has minimal vanishing at all cusps. $Gamma_0(4)$ is conjugate to $Gamma(2)$ by $tau mapsto 2tau$, so it suffices to check that $Delta(tau)$, the square of $eta(tau)^{12}$, vanishes to twice the minimum order at each cusp of $Gamma(2)$. The quotient $Gamma(1)/Gamma(2) cong S_3$ acts transitively on the cusps of $X(2)$ with stabilizers of order 2, so the quotient map to $X(1)$ has ramification degree 2 at each cusp. $Delta(tau)$ is invariant under the weight 12 action of $Gamma(1)$, and $Delta(tau)$ has minimal vanishing at infinity on $X(1)$.

Old answer: If you have a cusp form of weight $k/2$ for $Gamma_0(4)$ (e.g., given to you by SAGE), you can multiply it by the modular function $frac{eta(tau)^8}{eta(4tau)^8} = q^{-1} - 8 + 20q - 62q^3 + 216q^5 - dots$ to get a modular form of the same weight, that is nonvanishing at one of the three cusps and vanishing at the other two. If you want a form that is nonzero at one of the other cusps, you can multiply by $frac{eta(4tau)^8}{eta(tau)^8}$ (has a pole at zero) or by $frac{eta(tau)^{16}eta(4tau)^8}{eta(2tau)^{24}}$ (pole at $1/2$). [Constant term $-8$ added Sept. 23, in response to an email correction from Michael Somos.]