Edit: (1) below is now clear, but I'm still looking for an answer for (2).
There are two issues I have when trying to understand the classifying spaces of (oriented) vector bundles:
(1) Let $E \to X$ be an $n$-dim vector bundle with classifying map $f: X \to BO(n)$. The space $BSO(n)$ is the classifying space of oriented vector bundles and at the same time the universal cover of $BO(n)$. Then $E$ is orientable if and only if $f$ lifts to a map $\tilde{f}: X \to BSO(n)$.
Question 1: Why doesn't it contradict the fact that $f$ admits a lift if and only if $f_* (\pi_1 (X,x_0)) \subseteq p_* (\pi_1 (BSO(n),y_0))=0$? This is the lifting criterion for covering spaces, and since $BSO(n)$ is simply connected then for a lift to exist $X$ should be simply connected as well, but this is definitely not the case.
(2) I'm trying to show that $BSO(n)$ is simply connected without using that $\pi_1 (BSO(n)) = \pi_0 (SO(n))=0$, but rather using that $BSO(n)$ is the classifying space for oriented $n$-dimensional real vector bundles. Since there is a unique vector bundle over the one-point space, it is easy to see that $BSO(n)$ is path-connected.
However if I want to extend the argument for $\pi_1$, I bump into the following issue regarding basepoints:
$$ \{ S^1 \times \mathbb{R}^n \} = \{ n-\text{dim oriented vector bundles over }S^1 \} = [ S^1, BSO(n) ] \leftarrow \pi_1 (BSO(n)) $$
Question 2: The last map is surjective since $BSO(n)$ is path-connected, but how to show that it is bijective? This is true since $BSO(n)$ is simply connected, but I don't know that a priori (that's exactly what I am trying to show).
This approach is hinted in Hatcher's Vector bundles and $K$-theory, page 31, but without further comment.