Regarding your question on De Rham cohomology there are several approches to realize a De Rham complex that computes singular cohomology.
A. In Algebraic geometry.
You should look at R. Hartshorne "Algebraic De Rham cohomology" manuscripta math. 7, 125-140 (1972). It is a research announcement and survey on the cohomology of algebraic De Rham forms on algebraic varieties, details are published in the Publications of IHES (1975).
In particular for a scheme $Y$ of finite type over a characteristic zero field $k$, he defines its algebraic De Rham cohomology.
1) You embed $Y$ as a closed subscheme of a smooth scheme $X$.
2) You consider $\Omega^*$ the complex of sheaves of regular differential forms on $X$ over $k$.
3) You take $\hat{X}$ the formal completion of $X$ along $Y$:
$$\hat{Y}=\bigcup_n Y(n)$$
where $Y(n)$ is the infinitesimal neighbourhood of order $n$ of $Y$ in $X$
and consider $\hat{\Omega}^*$ the completion of $\Omega^*$.
4) You define $H^*_{DR}(Y)$ as the hypercohomology of the complex $\hat{\Omega}^*$ on the formal scheme $\hat{X}$.
Then (theorem 1.6 of this paper) when $Y$ is a scheme of finite type over $k=\mathbb{C}$ we have a natural isomorphism
$$H^i_{DR}(Y)\cong H^i(Y^{an},\mathbb{C})$$
where $Y^{an}$ is the corresponding complex analytic space and $H^i(-,\mathbb{C})$ is the singular cohomology.
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B. As stratified spaces
You can use the fact that a complex algebraic variety is stratified, for example it is a stratifold in the sense of M. Kreck, then you have a notion of De Rham complex that computes singular cohomology with real coefficients:
Or you can use Whitney functions
