I can construct the classifying space for real vector bundles as the Grassmanian $Gr(n, \mathbb R^\infty)$, and I can construct $BO(n)$ as well. They happen to be the same. But is there an easier way of seeing this, perhaps by noting something about transition functions of real vector bundles?
The same question applies for complex bundles and their classification space.
Yes, if $E/M$ is a vector bundle of rank $k$, you can embed that bundle inside the trivial bundle $M \times \Bbb R^n$ over $M$ for large enough $n$, and that gives a map $M \to \text{Gr}(k, n)$ by sending each point $p$ in $M$ to the embedding $\Bbb R^k \hookrightarrow \Bbb R^n$ (a $k$-dimensional subspace of $\Bbb R^n$) induced from the embedding $E \hookrightarrow M \times \Bbb R^n$. By varying $n$, you do get a map $M \to \text{Gr}(k, \infty)$ classifying $E/M$.
There are two ways to see that vector bundles of rank $k$ are also classified by maps $M \to BO(k)$: one is to directly note that $\text{Gr}(k, n)$ is a quotient of the Stiefel manifold, $V(k, n)$, consisting of all the orthonormal $k$-frames in $\Bbb R^n$. The map $p : V(k, n) \to \text{Gr}(k, n)$ sending a $k$-frame to the $k$-dimensional subspace spanned by them, is a principal bundle with fibers being acted upon freely and transitively by $O(k)$. Since for $n = \infty$ (direct limit), $V(k, \infty)$ is weakly contractible, we conclude $\text{Gr}(k, \infty)$ is a $BO(k)$ by definition.
On the other hand, you could understand that vector bundles of rank $k$ correspond bijectively to principal $O(k)$-bundles; take an arbitrary vector bundle $E/M$, reduce the structure group to $O(k) \subset GL_k(\Bbb R)$ by imposing a Riemannian metric, so that the cocycles $\varphi_{ij} : U_i \cap U_j \to O(k)$ are $O(k)$-valued. Construct a new principal bundle out of $\bigsqcup U_i \times O(k)$ with $U_i$ running over the bundle charts, and by identifying $(x, v) = (x, \varphi v)$ on overlaps. This is exactly the orthonormal frame bundle of $E$, which is a principal $O(k)$-bundle. Now principal $G$-bundles are classified by maps to $BG$, so vector bundles $E/M$ of rank $k$ are classified by the classifying map $M \to BO(k)$ of the orthonormal frame bundle of $E$.