Dear Victoria: here is a summary of the main comparison results I know of between Grothendieck cohomology (which is usually just called cohomology and written $\newcommand{\F}{\mathcal F}H^i(X,\F)$ ) and other cohomologies.
1) If $X$ is locally contractible then the cohomology of a constant sheaf coincides with singular cohomology. [This is Eric's answer, but there is no need for his hypothesis that open subsets be acyclic]
2) Cartan's theorem: Given a topological space $X$ and a sheaf $\F$, assume there exists a basis of open sets $\mathcal{U}$, stable under finite intersections, such that the CECH cohomology groups for the sheaf $\F$ are trivial (in positive dimension) for every open $U$ in the basis: $H^i(U,\F)=0$
Then the Cech cohomology of $\F$ on $X$ coincides with (Grothendieck) cohomology
3) Leray's Theorem: Given a topological space $X$ and a sheaf $\F$, assume that for some covering $(U_i)$ of $X$ we know that the (Grothendieck!) cohomology in positive dimensions of $\F$ vanishes on every finite intersection of the $U_i$'s.
Then the cohomology of $\F$ is already calculated by the Cech cohomology OF THE COVERING $(U_i)$: no need to pass to the inductive limit on all covers.
This contains Dinakar's favourite example of a quasi-coherent sheaf on a separated scheme covered by affines.
4) If $X$ is paracompact and Hausdorff, Cech cohomology coincides with Grothendieck cohomology for ALL SHEAVES
If you think this is too nice to be true, you can check Théorème 5.10.1 in Godement's book cited below
[So Eric's remark that no matter how nice the space is, Cech cohomology would probably not coincide with derived functor cohomology for arbitrary sheaves turns out to be too pessimistic]
5) Cohomology can be calculated by taking sections of any acyclic resolution of the studied sheaf: you don't need to take an injective resolution. This contains De Rham's theorem that singular cohomology can be calculated with differential forms on manifolds.
6) If you study sheaves of non-abelian groups, Cech cohomology is convenient: for example vector bundles on $X$ ( a topological space or manifold or scheme or...) are parametrized by $H^1(X, GL_r)$. I don't know if there is a description of sheaf cohomology for non-abelian sheaves in the derived functor style.
Good references are
a) A classic: Godement, Théorie des faisceaux (in French, alas)
b) S.Ramanan, Global Calculus,AMS graduate Studies in Mahematics, volume 65.
(An amazingly lucid book, in the best Indian tradition.)
c) Torsten Wedhorn's quite detailed on-line notes, which prove 1) above (Theorem 9.16, p.92) and much, much more.
By the way, @Wedhorn is one of the two authors of a great book on algebraic geometry.
d) Ciboratu, Proposition 2.1 and Voisin's Hodge Theory and Complex Algebraic Geometry I, Theorem 4.47, page 109 , which both also prove 1) above.
For Question #2: Look at chapter 2 of Bott, R. & Lu, T. "Differential forms in Algebraic Topology", Springer GTM vol 82. There you will find a nice introduction to Cech cohomology through a Mayer-Vietoris general point of view.
For Question #4: Yes. Look at Bredon, G. "Topology and Geometry", Springer GTM vol 139, p. 289. There is a Mayer-Vietoris set-construction which is a nice way to avoid presheaf calculations.
Best Answer
For the first question: The higher cohomology of a paracompact Hausdorff space (e.g. manifold) in the sheaf of continuous real valued functions is zero, because the sheaf is fine.
If I understand the spirit of your second question correctly, then you can use locally constant sheaves or more generally constructible sheaves to get interesting "topological" cohomologies.
A quick follow up: (1) The argument in the first paragraph also applies to sheaves of modules over the ring of continuous functions. (2) I don't know if every generalized cohomology theory (in the sense of algebraic topology) can or should be regarded as sheaf cohomology, but I leave to an expert to give a precise answer. On the flip side, I should point out that sheaf cohomology with arbitrary coefficients is generally not homotopy invariant, so it really is a different sort of beast.