Why is the one-point compactification of $S^5\times \mathbb{R}$ not a manifold?

First, note that it's not sufficient to prove that $X$ is homotopy equivalent to $S^6 \vee S^1$ in order to prove that it's not a manifold. Indeed you could "thicken" $S^6 \vee S^1$ and get an 8-manifold (if you want, imagine figure eight $S^1 \vee S^1$ embedded in the plane $R^2$, and consider a small neighborhood of that: that's a 2-manifold; the idea is the same). "Being a manifold" is not preserved by homotopy equivalence.

Now suppose that $X = S^5 \times \mathbb{R} \cup \{ \infty \}$ were an $n$-manifold. Then by definition, $\infty \in X$ must have a neighborhood $U \subset X$ homeomorphic to $D^n$, where $\infty$ corresponds to $0 \in D^n$. Using excision, you get $$H_k(X, S^5 \times \mathbb{R}) \cong H_k(U, U \setminus \{ \infty \})$$ and since homology is invariant under homeomorphisms, you get: $$H_k(X, S^5 \times \mathbb{R} \cup \{ \infty \}) \cong H_k(D^n, D^n \setminus \{0\}) \cong \begin{cases} \mathbb{Z} & \text{if } k = n; \\ 0 & \text{if } k \neq n. \end{cases}$$

Now, clearly any point in $X \setminus \{\infty\}$ has a neighborhood homeomorphic to $D^6$ (because $S^5 \times \mathbb{R}$ is a $6$-manifold), thus $n = 6$. If you write down the long exact sequence in homology of the pair $(X, S^5 \times \mathbb{R})$, you get: $$H_{k-1}(X, S^5 \times \mathbb{R}) \to H_k(S^5) \to H_k(X) \to H_k(X, S^5 \times \mathbb{R}) \to H_{k+1}(S^5 \times \mathbb{R})$$ The homology of $S^5 \times \mathbb{R} \simeq S^5$ is well-known. We thus get that $$H_k(X) = \begin{cases} \mathbb{Z} & \text{if } k = 0,5,6; \\ 0 & \text{otherwise.} \end{cases}$$

From $H_6(X) \cong \mathbb{Z}$ we deduce that $X$ is orientable. It must therefore satisfy Poincaré duality, in particular $H^1(X) \cong H_5(X) \cong \mathbb{Z}$. But by the universal coefficients theorem, $H^1(X) \cong \hom_\mathbb{Z}(H_1(X), \mathbb{Z}) = 0$. This is a contradiction.