Klein bottle in $\mathbb{R}^4$ does not have a couple of normal vector fields

By a couple of normal vector fields, I assume you mean two linearly independent normal vector fields.

Let $i : K \to \mathbb{R}^4$ be an embedding. On $K$ we have an exact sequence of vector bundles

$$0 \to TK \to i^*T\mathbb{R}^4 \to N \to 0$$

where $N$ is the normal bundle; note that $\operatorname{rank}N = \dim\mathbb{R}^4 - \dim K = 4 - 2 = 2$. If $K$ had two linearly independent normal vector fields, $N$ would be trivial. Applying the first Stiefel-Whitney class, we see that $w_1(i^*T\mathbb{R}^4) = w_1(TK) + w_1(N)$, but as $T\mathbb{R}^4$ and $N$ are both trivial, $w_1(i^*T\mathbb{R}^4) = i^*w_1(T\mathbb{R}^4) = i^*0 = 0$ and $w_1(N) = 0$, so $w_1(TK) = 0$. But this is impossible as the first Stiefel-Whitney class of the tangent bundle vanishes if and only if the manifold is orientable, which $K$ is not.

In general, a closed manifold $M$ admits an embedding into $\mathbb{R}^n$ with trivial normal bundle if and only if the manifold is stably parallelisable, i.e. $TM\oplus\varepsilon^k$ is trivial for some $k$. Therefore, if such an embedding exists, all the Stiefel-Whitney classes of $M$ must vanish; in particular, we see that $M$ must be orientable.

Came up with a bit simpler solution. Let $e_1(x)$ be a tangent vector field on the Klein bottle $K$. Suppose that $n_1(x)$ and $n_2(x)$ - pair of linearly indenpendent normal vector fields. Then we can choose $e_2(x)$ in the $TK_x$ so that four vectors $(e_1(x), e_2(x), n_1(x), n_2(x))$ form a positively oriented basis in $\mathbb{R}^4$. This would give us orientation on the Klein bottle.