Manifold non-orientable iff. frame bundle is connected

Let $M$ be a connected smooth manifold and $L(M):=\bigcup_{x\in M}L_xM$ its frame bundle where $L_xM:=\{(v_1,\dots,v_n):\{v_1,\dots,v_n\}\text{ is a basis of }T_xM\}$.

$M$ is non-orientable iff. $L(M)$ is connected.

I think I have managed to prove, that if $L(M)$ is connected then $M$ is non-orientable by assuming $M$ is orientable and showing that then $L(M)$ has two disjoint components, namely the set of all positively oriented bases and the "rest". However, I struggle on proving the converse. My idea is to use non-orientability to show path-connectedness but I cannot construct such a continuous path in $L(M)$.

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It's a fact that given a path $c:[0,1] \to M$ and a (positive) basis of $T_{c(0)} M$, you can "move" this basis to $c(1)$ (the formal definition would be $\exists$ vector fields $v_1,\dots,v_n$, where $n = \dim M$, along the path such that $\{v_i(0)\}$ coincides with your starting basis), by continuity the basis of $T_{c(1)} M$ is also positive.
Now try to prove that $M$ is orientable iff for every loop $\lambda:[0,1] \to M$, the orientation on $T_{\lambda(0)}M$ is preserved when you move it to $T_{\lambda(1)}M$ in the above sense. Can you see how your question follows?
Since $M$ is non-orientable, we can find for any $p\in M$ a loop $\lambda$ s.t. the orientation along $\lambda$ is reversed. Since $M$ is also connected, for arbitrary $p,q\in M$ we can find a continuous path $\gamma$ s.t. $\gamma(0)=p$, $\gamma(1)=q$. Then, by composing $\lambda\circ\gamma$ if necessary (i.e. if $\{v_i(p)\}$ and $\{v_i(q)\}$ determine opposite orientations of $T_pM$ and $T_qM$, respectively) we can connect any two bases in $L(M)$.