Given an equation in Sturm Liouville form, how does one write it in canonical form?
For instance, the Sturm Liouville equation $(2\sqrt{t}x')'+\frac{\lambda}{\sqrt{t}}x=0, x'(1)=0, x'(4)=0, t>0, \lambda>0$ can be written in canonical form as $4\sqrt{t}x''+\frac{2}{\sqrt{t}}x'+\frac{\lambda}{\sqrt{t}}x=0$.
In your example, I think you mean
\begin{align} \ & \frac{d}{dt}\biggl (2\sqrt t \frac{dx}{dt} \biggl)+\frac{\lambda}{\sqrt t}x=0 \\ \ \implies & 2\sqrt t\frac{d}{dt}\biggl (\frac{dx}{dt}\biggl)+\frac{dx}{dt}\Bigl (2\sqrt t \Bigl)+\frac{\lambda}{\sqrt t}x=0 \\ \ \implies & 2\sqrt t\frac{d^2x}{dt^2}+\frac{1}{\sqrt t}\frac{dx}{dt}+\frac{\lambda}{\sqrt t}x=0 \\ \ \implies & 4\sqrt tx''+\frac{2}{\sqrt t}x'+\frac{2\lambda}{\sqrt t}x=0 \end{align}
If this is what you mean by converting into canonical form, then it really is just a matter of computing the derivatives:
\begin{align} \ & \frac{d}{dt}\biggl (p(t) \frac{dx}{dt} \biggl )+q(t)x=0 \\ \ \implies & p(t)\frac{d}{dt}\biggl (\frac{dx}{dt} \biggl )+\frac{dx}{dt}\frac{d}{dt}\bigl (p(t) \bigl )+q(t)x=0 \\ \ \implies & p(t)x''+p'(t)x'+q(t)x=0 \end{align}
The other direction is perhaps more interesting:
\begin{align} \ & p(t)x''+q(t)x'+r(t)x=0 \\ \ \implies & \frac{d^2x}{dt^2}+\frac{q(t)}{p(t)} \frac{dx}{dt}+\frac{r(t)}{p(t)}x=0 \\ \ \implies & I(t)\frac{d^2x}{dt^2}+I(t)\frac{q(t)}{p(t)} \frac{dx}{dt}+I(t)\frac{r(t)}{p(t)}x=0 && \text{where }I(t):=\exp \biggl(\int \frac{q(t)}{p(t)}dt \biggr)\\ \ \implies & \frac{d}{dt}\biggl(I(t)\frac{dx}{dt} \biggl)+I(t)\frac{r(t)}{p(t)}x=0 \end{align}
Analogous to dealing with equations of the form $y'+f(x)y=g(x)$, $I(t)$ here is like the "integrating factor".