It does matter what kind of cohomology you use:
First think of the connected sum of an infinite number of real projective spaces end to end. De Rham cohomology cannot tell that these were not spheres, but mod 2 cohomology shows that the thing is not of finite type.
Now take it further. Take a (necessarily non simply connected) closed manifold whose integral homology is the same as that of a sphere (for example Poincare's 3-dimensional example whose fundamental group is the binary icosahedral group). Just as this looks like sphere to integral cohomology, the connected sum of infinitely many of these appears is indistinguishable from the connected sum of infinitely many spheres.
If you allow cohomology with twisted coefficients, then things get more hopeful. I would be inclined to separate the problem into two parts: (1) if a smooth manifold has the homotopy type of a finite CW complex, does this imply that it is of finite type in your sense? (2) What finiteness conditions on cohomology suffice to guarantee that a CW complex (space) has the (weak) homotopy type of a finite CW complex? (2) has the expected affirmative answer if you confine yourself to simply connected spaces, but the general case is more subtle.
EDIT What I really mean is: First assume that there are finitely many components, and that each of these has a finite presented fundamental group. Then use Wall's finiteness obstruction (which has to do with modules over the group ring of the fund gp -- it's a long story).
RE-EDIT Some more details: There are finite groups $G$ for which the ring $\mathbb ZG$ has some projective modules that are not stably free. For any such $G$ there are examples of finite CW complexes $K$ and $L$ with fundamental group $G$ such that $L$ is finite and $K$ is a retract of $L$ up to homotopy (i.e. there are maps $K\to L\to K$ whose composition is homotopic to the identity) but $K$ is not homotopy equivalent to any finite complex. The other composition $p:L\to L$ is then such that $p\circ p$ is homotopic to $p$, and $K$ is equivalent to the mapping telescope or homotopy colimit of $L\to L\to L\to\dots$. Take any such example and realize it by manifolds: Choose $L$ to be a smooth compact parallelizable manifold with boundary and (by taking the dimension big enough) take $p$ to be a smooth embedding of $L$ in the interior of $L$, and take $K$ to be the increasing union of infinitely many nested copies of $L$. So $K$ is a noncompact manifold without boundary whose (co)homology is in any sense no bigger than that of $L$, but which is not homotopy equivalent to any finite complex. In particular it is not the interior of any compact manifold with boundary. I can't see why it cannot be of finite type in the sense of the problem, but it looks unlikely.
It just struck me that this example can be improved: By doing surgery on a finite number of embedded $1$-spheres, one can arrange for the noncompact manifold $K$ to be simply connected. In this case it is homotopy equivalent to a finite complex (basically because $\mathbb Z$-modules are (stably) free), but it still is not the interior of any smooth compact manifold (because outside a compact set it's still the same example).
Background to this (but I am not using the full strength of it) is Wall's finiteness obstruction for CW complexes and Siebenmann's thesis in which Wall's obstruction is adapted to give an obstruction to putting a boundary on an isolated end of a manifold.
I am not aware of a general result regarding the existence of good covers (and would guess that the general answer is negative). However, if you are willing to make certain sacrifices in terms of additional hypothesis and weaker conclusions, then you may seek solace in the article
RP Osborne and JL Stern. Covering Manifolds with Cells, Pacific Journal of Mathematics, Vol 30, No. 1, 1969.
And here, for ready reference, is the main theorem.
Let $M$ be a $k$-connected topological $n$-manifold without boundary, and define $q = \min(k,n-3)$. $M$ admits a cover by $p$ open cells if $p(q+1) > n$ and moreover, one can arrange that all nonempty intersections of the covering cells are $(q-1)$-connected.
What's nice here is that one actually gets a bound on the size of the open cover, but of course the intersections need not be contractible and you require many homotopy groups of the original manifold to vanish.
Best Answer
you don't really need a whole lot of Riemannian geometry to prove this. Embed the manifold into $\mathbb R^n$ by Whitney and look at very small charts around points given by orthogonal projections onto the tangent spaces. the transition maps will be arbitrary close to identity in $C^2$. that means that a small round disk in one chart will remain strictly convex in nearby charts (because if $f(x)=|x|^2$ and $\phi$ is a transition map such that $\phi-Id$ has small first and second derivatives then $f\circ \phi$ is still strictly convex and hence has convex sublevel sets). This is is all you need to conclude that all intersections are contractible. I guess since the above argument doesn't use any Riemannian geometry notions it should qualify as an answer to the second question?
Incidentally, does a good open cover always exist if a manifold is only topological?