From the Leray-Serre spectral sequence for a covering map $Y\to X$, which is a fibration with discrete fibers, we get an isomorphism $H^p(Y)\cong H^p(X,\mathcal H^0)$, where $\mathcal H^0$ denotes the local system of coefficients which at each point of $X$ has group equal to $H^0(p^{-1}(x))$.
If the covering is of $n$ sheets, then $\mathcal H^0$ is locally $\mathbb Z^n$, with the fundamental group of $X$ acting by permutation of the standard basis according to the monodromy permutation representation.
Now the local system $\mathcal H^0$ corresponds to a sheaf $\mathcal F$ on $X$, and for sensible $X$ (paracompact, say), one can compute singular cohomology with coefficients in the local system as sheaf cohomology with coefficients on the sheaf $\mathcal F$. If $X$ has a good finite cover $\mathcal U$ (in the sense of the book of Bott-Tu) then one can compute sheaf cohomoogy $H^p(X,\mathcal H^0)$ as the Cech cohomology $H^p(\mathcal U,\mathcal H^0)$. Looking at the complex which computes this by definition, we see that the Euler characteristic of $H^p(\mathcal U,\mathcal H^0)$, and therefore of $H^\bullet(X,\mathcal H^0)$, is $n$ times that of $H^\bullet(X,\mathbb Z)$. Notice that the existence of good covers implies being of bounded finite type, as you say (but I think it even implies that the space of of the homotopy type of a CW-complex, namely the nerve of the good covering... so all this might not get us much)
The fact that the Euler characteristic of a sensible space with coefficients on a local system of coefficients which locally looks like $\mathbb Z^n$ is $n$ times that of the space should be written down somewhere, but I cannot find it now.
There is this answer by Matt but he does not give a reference.
We will prove this by induction: suppose, we know that if $X$ consists of $\,<n$ cells then for all other finite $CW$-complexes $X'$ such that $X\approx X'$ we have $\chi(X)=\chi(X')$.
Let $Y$ and $Y'$ be two finite $CW$-complexes, $Y$ consists of $n$ cells, and $f:Y\to Y'$ is homotopy equivalence. Consider one higher-dimensional cell of $Y$: let $Y=Z\cup_\alpha D^k$, where $Z$ consists of $n-1$ cells, $D^k$ is just a cell and $\alpha:\partial D^k\to Z$ is attaching map. We see that $\chi(Z)=\chi(Y)-(-1^k)$.
Then consider the space $CZ\cup_{f|_Z}Y'$, here $CZ$ is a cone. $f$ supposed to be cellular map, so $CZ\cup_{f|_Z}Y'$ is a $CW$ complex; it contains all cells of $Y'$, all cells of $Z$ times $I$, and the vertex of the cone, thus $\chi(CZ\cup_{f|_Z}Y')=\chi(Y')-\chi(Z)+1$. But we know that this space has homotopy type of $S^k$, so by inductive hypothesis $\chi(CZ\cup_{f|_Z}Y')=1+(-1^k)$, and $\chi(Y)=\chi(Y')$ as desired.
EDIT: for the induction we also need the statement be truth in case $X\approx pt$ and $X\approx S^m$. If $X\approx S^m$, we may glue an $m+1$-disc and obtain contractible space.
Now suppose $X\approx pt$, $m$ is maximal dimension of cells of $X$ and $X$ has $p$ $\,m$-cells. The space $sk_{m-1}(X)$ is $(m-2)$-connected, so it is homotopy equivalent to the bouquet of $q$ $\,(n-1)$-spheres. Gluing $m$-cells determines a homomorphism $\phi:\mathbb Z^p\to\mathbb Z^q$, and equalities $\mathrm{coim\,}\phi=\pi_{m-1}(X)$ and $\ker\phi\subseteq\pi_m(X)$ give us $p=q$. Then we may remove $p$ $\,m$-cells, $p$ $\,(m-1)$-cells, and repeat. (this reasoning sounds more like homological argument)
Best Answer
Not really. If you want to use the definition as the alternating sum of the number of cells, then it certainly won't work; the sum $\sum_k (-1)^k c_k$ will never converge if an infinite number of $c_k$ are nonzero. If $\operatorname{rank} H_k(X) = 0$ for $k \ge N$ (with a fixed $N$) then you can use the second definition. But if an infinite number of $H_k(X)$ have positive rank, then it won't work.
What can be done is defining the Poincaré polynomial of $X$. It's given by: $$P_X(t) = \sum_{k \ge 0} (\operatorname{rank} H_k(X)) t^k.$$ In general it's an infinite series. If the Euler characteristic is well-defined (i.e. if only a finite number of homology groups have positive ranks), then $\chi(X) = P_X(-1)$. Otherwise, you cannot say much more.