I eventually realised that in order to answer my question I needed to understand the derivation of D&L's Equation 25: $$H\left(z\right)=H_{0}\left(1+z\right)\left[1+\varOmega_{m}z+\varOmega_{\varLambda}\left(\frac{1}{\left(1+z\right)^{2}}-1\right)\right]^{1/2}.$$
This is the Hubble parameter parameterised by redshift not time. The critical density at present is$$\rho_{c}=\frac{3H_{0}^{2}}{8\pi G}.$$
Therefore $$H_{0}^{2}=\frac{8\pi G\rho_{c}}{3}$$
and $$\varOmega=\frac{\rho}{\rho_{c}}=\frac{8\pi G\rho}{3H_{0}^{2}}.$$
The Friedmann equation is
$$H^{2}\left(t\right)=\frac{8\pi G}{3}\rho-\frac{kc^{2}}{R^{2}}.$$
Divide by $H_{0}^{2}$ $$\frac{H^{2}}{H_{0}^{2}}=\frac{8\pi G}{3H_{0}^{2}}\rho-\frac{kc^{2}}{R^{2}H_{0}^{2}}$$
$$\frac{H^{2}}{H_{0}^{2}}=\varOmega-\frac{kc^{2}}{R^{2}H_{0}^{2}}.$$Where $\varOmega=\varOmega_{m}+\varOmega_{\varLambda}+\varOmega_{r}$. To find $-kc^{2}$ set$$-kc^{2}=H_{0}^{2}R_{0}^{2}-\frac{R_{0}^{2}8\pi G\rho}{3}$$ and get$$\frac{H^{2}}{H_{0}^{2}}=\varOmega+\frac{H_{0}^{2}R_{0}^{2}}{H_{0}^{2}R^{2}}-\frac{R_{0}^{2}8\pi G\rho}{3R^{2}H_{0}^{2}}$$
$$\frac{H^{2}}{H_{0}^{2}}=\varOmega+\frac{R_{0}^{2}}{R^{2}}\left(1-\frac{8\pi G\rho}{3H_{0}^{2}}\right)$$ $$\frac{H^{2}}{H_{0}^{2}}=\varOmega+\frac{R_{0}^{2}}{R^{2}}\left(1-\varOmega\right).$$
Now write as a function of redshift $z$.
First, $$\frac{R_{0}^{2}}{R^{2}}=\left(1+z\right)^{2}$$giving$$H^{2}\left(z\right)=H_{0}^{2}\left[\varOmega+\left(1+z\right)^{2}\left(1-\varOmega\right)\right].$$
And$$\rho\left(z\right)=\rho_{m}\left(1+z\right)^{3}+\rho_{\varLambda}+\rho_{r}\left(1+z\right)^{4}$$
giving$$H^{2}\left(z\right)=H_{0}^{2}\left[\varOmega_{m}\left(1+z\right)^{3}+\varOmega_{\varLambda}+\varOmega_{r}\left(1+z\right)^{4}+\left(1+z\right)^{2}\left(1-\varOmega\right)\right].$$To simplify, let $\varOmega_{r}=0$ (as do Davis and Lineweaver)$$H^{2}\left(z\right)=H_{0}^{2}\left[\varOmega_{m}\left(1+z\right)^{3}+\varOmega_{\varLambda}+\left(1+z\right)^{2}\left(1-\varOmega_{m}-\varOmega_{\varLambda}\right)\right]$$ $$H^{2}\left(z\right)=H_{0}^{2}\left(1+z\right)^{2}\left[\varOmega_{m}\left(1+z\right)+\frac{\varOmega_{\varLambda}}{\left(1+z\right)^{2}}+1-\varOmega_{m}-\varOmega_{\varLambda}\right]$$ $$H^{2}\left(z\right)=H_{0}^{2}\left(1+z\right)^{2}\left[1+\varOmega_{m}z+\varOmega_{\varLambda}\left(\frac{1}{\left(1+z\right)^{2}}-1\right)\right]$$
$$H\left(z\right)=H_{0}\left(1+z\right)\left[1+\varOmega_{m}z+\varOmega_{\varLambda}\left(\frac{1}{\left(1+z\right)^{2}}-1\right)\right]^{1/2}.$$