Let $E$ be a vector bundle of rank $k$ over a smooth manifold $M$ of dimension $m$. Then $E$ can be thought of as a $(m+k)$-dimensional manifold, so the (weak) Whitney immersion theorem states that $E$ can be immersed in $\mathbb{R}^{2(m+k)}$.
Is it possible to immerse $E$ in $\mathbb{R}^{2m + k}$?
Here is an argument filling in the sketch provided by @MikeMiller in the comments.
Fix an immersion $M \looparrowright \mathbb{R}^{2m}$ with normal bundle $\nu_M$, which exists by the weak Whitney immersion theorem. It suffices to show that there is a bundle monomorphism $E \hookrightarrow \nu_M \oplus \mathbb{R}^k$. The tubular neighborhood theorem asserts that there is a diffeomorphism of $\nu_M$ onto a subspace of $\mathbb{R}^{2m}$, and so this gives an immersion of $E$ into $\mathbb{R}^{2m} \times \mathbb{R}^k \cong \mathbb{R}^{2m + k}$.
It remains to show that such a bundle monomorphism exists. This follows from the more general proposition.
The proof of this proposition proceeds by obstruction theory. The given bundles $E$ and $G$ furnish a classifying map $M \to BO(k) \times BO(k+m)$. In terms of classifying maps, the conclusion of the proposition is equivalent to showing that the map $M \to BO(k) \times BO(k+m)$ lifts to $BO(k) \times BO(m)$: $$ \require{AMScd} \begin{CD} @. BO(k) \times BO(m) \\ {} @V{1 \times \mu}VV \\ M @>>> BO(k) \times BO(k+m) \end{CD}$$
Note that the cofiber of the vertical map has no positive-dimensional cells in dimensions $\leq m$, while $M$ can be approximated by a $m$-dimensional cell complex. So the map from $M$ to the cofiber is null, and hence a lift exists.