The results you need in order to prove that the Klein bottle can't be embedded in $\mathbb{R^3}$:
$\bullet$ The Klein bottle isn't orientable.
$\bullet$ The Klein bottle is a compact, and connected 2-manifold.
$\bullet$ Any compact, connected manifold without boundary of dimension $n$ embedded in $\mathbb{R^{n+1}}$ is orientable.
Therefore, the Klein bottle can't be embedded in $\mathbb{R^3}$.
It seems like you might be confused regarding well-definedness of normal bundles and well-definedness of fiber dimension. So I will write an answer focussing on that issue.
The normal bundle of an $m$-dimensional manifold $M$ is not a well-defined thing. First you have to choose an embedding (or an immersion) of $M$: only then can you talk about the normal bundle of $M$ relative to that embedding (or immersion); and only then can you talk about the dimension of the fibers of that normal bundle.
Maybe $M$ was already given to you as a submanifold of $\mathbb R^n$ so you can certainly talk about the normal bundle of the inclusion map (which is an embedding), but that should not deter you from studying other (noninclusion) embeddings of $M$ and their normal bundles.
For example, if you embed the circle $S^1$ into the plane $\mathbb R^2$ then the normal bundle of that embedding has fibers of dimension $2-1=1$, i.e. lines. But if on the other hand you embed $S^1$ as a knot in $\mathbb R^3$ then the normal bundle of this new embedding has fibers of dimension $3-1=2$, i.e. planes.
Regarding "the" Klein bottle $K$ (scare quotes intentional), which is a 2-dimensional manifold, like any manifold it has many different embeddings. The lowest dimensional Euclidean space into which $K$ embeds is $\mathbb R^4$, and under any such embedding the normal bundle has fibers of dimension $4-2=2$. In fact you can embed $K$ into Euclidean space $\mathbb R^n$ of any dimension $n \ge 4$, and you would get a normal bundle with fibers of dimension $n-2$.
The usual image of a Klein bottle in $\mathbb R^3$ is not an embedding but it is an immersion, and one can still define the normal bundle of an immersion. And the same formula holds: the normal bundle of an $m$-dimensional manifold immersed into $\mathbb R^n$ has fibers of dimension $n-m$. So the normal bundle of a Klein bottle immersed into $\mathbb R^3$ has fibers of dimension $3-2=1$, i.e. lines.
Best Answer
Understandably there are a lot of answers, but if you still have any further questions maybe this will help.
An embedding of a topological space $X$ into a topological space $Y$ is a continuous map $e \colon X \to Y$ such that $e|_X$ is a homeomorphism onto its image.
Both the Klein bottle ($f \colon I^2 / \sim \to \mathbb{R}^3$) is not embedded into $\mathbb{R}^3$, because it has self-intersections; this means that the immersion of the Klein bottle is not a bijection, hence not a homeomorphism, so not an embedding.
As I understand it, an immersion simply means that the tangent spaces are mapped injectively; i.e. that the map $D_p f \colon T_pI^2 \to T_{f(p)}\mathbb{R}^3$ is injective. In the Klein bottle example, at the self-intersection, any point of intersection has two distinct tangent planes, hence this map is injective.
I hope this makes some sense!