1) Every compact topological manifold is homotopy equivalent to a finite CW complex; see here. As a (very sketchy) sketch: You know via Hatcher that they're dominated by a finite CW complex, hence you can apply Wall's obstruction theory to being homotopy equivalent to a finite CW complex: see here. This immediately implies that every simply connected compact manifold is homotopy equivalent to a finite CW complex, and with more difficulty, that this is true for manifolds with eg fundamental group $\Bbb Z^n$. The general case is incredibly hard; the reference given in the MathOverflow post is as far as I know essentially the only proof of this fact. I would only suggest reading if you've got sufficient spunk. (For reference: I don't.)
2) Every compact smooth manifold is homeomorphic to a finite CW complex. This follows from Morse theory, which on a smooth manifold actually gives you a triangulation. This is much more elementary than (1).
3) That'd be Wall's finiteness obstruction, mentioned in the first paragraph. The linked notes of Lurie are accessible given a first course in algebraic topology, some experience with homological algebra, and some patience. To use it, you need your spaces to be finitely dominated; this is equivalent to satisfying Lurie's Lemma 6, which is not particularly helpful in practice. Usually you'll start with a space you actually know is finitely dominated and start using the finiteness obstruction then.
If you're willing to possibly care about countable CW complexes, this brief article of Milnor's implies every mapping space of finite CW complexes has the homotopy type of a countable CW complex.
Yes, you get a deformation retract. More formally...
Suppose $M$ is a topological manifold, $N\subseteq \partial M \subseteq M$ is a component of $\partial M$ and that the inclusion $N\rightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.
Here's the idea of the proof:
From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.
Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.
So, some closed neighborhood of $N$ in $M$ is of the form $N\times [0,1] = N\times I$, where we are identifying $N$ with $N\times \{0\}\subseteq M$..
Now suppose $f:M\rightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:N\times I\rightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:M\times I\rightarrow X$ satisfying two conditions. First, for any $n\in N$, $t\in I$, that $G(n,t) = F(n,t)$. Second, for any $m\in M$, $G(m,1) = f(m)$.
To that end, set $G(m,t) = \begin{cases} f(m) & m\notin N\times [0,1]\\ F(n, \max\{s,t\}) & m=(n,s)\in N\times [0,1].\end{cases}$
To check the first condition, note that $n\in N\cong N\times \{0\}$ means the corresonding $s$ is $s= 0$. Since $t\in[0,1]$, $\max\{s,t\} = \max\{0,t\}= t$, so $G(n,t) = F(n,t).$
To check the second condition, we have $G(m,t) = f(m)$ if $m\notin N\times [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)\in N\times [0,1]$, then because we are taking $t=1$, we have $\max\{s,t\} = t$. So $G(m,1) = F(m,1) = f(m)$.
Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $\max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.
But, if $m\in N\times [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $\max\{s,t\} = s=1$, so $G(m) = F(n, \max\{s,t\}) = F(n,1) = f(n).$
Best Answer
Every topological manifold is homotopy-equivalent to a finite-dimensional locally-finite simplicial complex. (This should be in Hatcher's book, see also here and here.) Hence, the wedge sum of topological manifolds is also homotopy-equivalent to a finite-dimensional locally-finite simplicial complex. Lastly, embedding such a complex as a subcomplex in some ${\mathbb R}^N$ and taking an open regular neighborhood of the image yields a smooth open manifold homotopy-equivalent to the above wedge sum.
Of course, if you are insisting on closed manifolds, then the answer is quite different: if $M_1, M_2$ are closed connected $n$-dimensional manifolds ($n\ge 1$), then $H_n(M_1\vee M_2, {\mathbb Z}_2)\cong {\mathbb Z}_2^2$, which implies that $M_1\vee M_2$ cannot be homotopy-equivalent to a closed connected $n$-dimensional manifold. One sees that disconnected manifolds cannot be homotopy-equivalent to $M_1\vee M_2$ by looking at $H_0$. One similarly rules out closed manifolds of other dimensions.