I have a suggestion for your problem.
The idea is due, to my knowledge, to Hassler Whitney from the 1940's, and it was used in a paper of William Massey from the 1960's about an estimate on the signature of a closed four manifold constructed from a surface in $S^4$.
Whitney observed an embedded closed surface in four space has a well-defined intersection number that does not require a global orientation for the surface.
Choose a local orientation near each point of transversal intersection of the surface with a pushed off copy of itself.
Both branches now have orientations which give a local orientation of four space which can be compared to a background orientation of four space. This gives a well-defined sign to the intersection point, note the order of branches does not affect the sign because they are even dimensional.
Changing the local orientation on the surface changes the orientation on each branch, so two minus signs.
It is a subtle and beautiful construction.
I am thinking if the algebraic sum of these is nonzero then it obstructs embedding a cone, because from a cone embedding one would get a embedding of the surface $\times$ $\epsilon$-interval, which would move the surface off of itself locally, forcing the Whitney sum to be zero.
The following lemma is an analogue of Corollary 3.46 in Hatcher’s “Algebraic Topology” and is proven by exactly the same method, using Alexander Duality (using cohomology with compact support instead of the ordinary cohomology as Hatcher does):
Lemma. Let $X$ be a locally contractible closed subset of $R^n$. Then $H_c^{n-2}(X)$ is torsion-free.
Corollary. If $X$ is a nonorientable connected pseudomanifold of dimension $n-1$, it cannot be properly embedded in $R^n$.
Proof. Such $X$ will have $2$-torsion in $H_c^{n-2}(X,{\mathbb Z})$. qed
Now, turning to the open cone $CK$ over the Klein bottle, i.e.
$(K\times [0,1))/(K\times 0)$. This space is nonorientable connected pseudomanifold. Each open subset of $CK$ containing the apex $a$ of this cone (the projection of $K\times 0$) is also a nonorientable pseudomanifold.
Suppose that $CK$ embeds in $R^4$. Take the intersection $Y$ of $CK$ with a sufficiently small open ball $B(a,r)$ in $R^4$. Use the homeomorphism $B(a,r)\to R^4$ to identify $Y$ with a closed subset $X\subset R^4$. We obtain a contradiction with Corollary above. Hence $CK$ does not embed in $R^4$.
Note that this argument is much more general applies to cones over compact nonorientable manifolds of any dimension.
Edit. Here is a better argument replacing Lemma and Corollary. I will use functoriality of the Alexander duality isomorphism AD: Given any closed subset $X\subset R^n$, we have a commutative square
$$
\begin{array}{ccc}
\check{H}^{n-1}_c(X; {\mathbb Z}) &\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}) \\
\alpha\downarrow & & \beta\downarrow \\
\check{H}^{n-1}_c(X; {\mathbb Z}/2)&\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}/2)
\end{array}
$$
Here cohomology is the compactly supported Chech cohomology. In the case when $X$ is locally contractible, it is naturally isomorphic to the compactly supported singular cohomology.
Now, if $X$ is homeomorphic to a nonorientable connected pseudomanifold of dimension $n-1$, then the homomorphism $\alpha$ is not surjective (since it is a homomorphism $0\to {\mathbb Z}/2$). On the other hand, the homomorphism $\beta$ is always surjective (as a homomorphism of $0$-degree reduced homology groups). Therefore, a nonorientable connected pseudomanifold of dimension $n-1$ cannot be properly topologically embedded in $R^n$. The rest of the argument is as above.
The cone on a manifold is not really a closed psuedomanifold.
Can you describe the embedding of the Klein bottle $\times$ interval?
With regards to the "better argument" that has been written, I like this proof. It shows the intersection with the link argument disproving nice embeddings, also works in spirit for all topological embeddings using cohomology etc.
It also shows $X$ locally separates $\mathbb{R}^n$, which is of course intuitive.
