Let $M$ be a closed $n$-dimensional smooth manifold. Let $X = \amalg_{p \in M} \text{Aut}(T_p M)$ be a fiber bundle over $M$ topologized in the natural way. It can be shown that the space of sections of $X$, $\Gamma(X)$, is isomorphic to $\text{Aut}(TM)$, the self-automorphisms fixing each tangent space $T_p M$.
Let $C(M, GL_n(\mathbb{R}))$ be the space of continuous maps from $M$ to $GL_n(\mathbb{R})$. Fix a basis $B$ of $\mathbb{R}^n$. I want to consider the following map, and determine whether it is well-defined.
Let $g: \Gamma(X) \to C(M, GL_n (\mathbb{R}))$ be such that it sends a section $\sigma$ to the map $\varphi$ where the linear automorphism $\sigma(p) \in \text{Aut}(T_p M)$ is sent to $\varphi(p)$, the unique real $n \times n$ matrix associated to it with respect to $B$. Does this make sense? I further claim, if this makes sense, that it would be an injective homomorphism.
I was told by my supervisor that there is a problem and that $g$ is well defined only if $M$ is parallelizable, because we need a continuous choice of basis of $T_pM$ for each $p$. (Then $g$ would in fact be an isomorphism.) I thought I understood this reasoning, but thinking back it seems to me that the basis of each tangent space doesn't matter, since that has no effect on the linear automorphism $\sigma(p)$ or its corresponding matrix, and only $B$ matters. Is this true, or am I missing some key linear algebra concept?
This map is not well-defined. To define it you need to pick, continuously, a collection of isomorphisms $GL_n(\mathbb{R}) \cong \text{Aut}(T_p(M))$, and you aren't guaranteed to be able to do this. There is no "unique real $n \times n$ matrix" as you claim. To write down such a map it suffices that there exists a trivialization of $T \otimes T^{\ast} \cong \text{End}(T)$ (and there need not exist such a trivialization) as Kajelad notes in the comments; I'm not sure if this is necessary.
I think this would follow if the smooth automorphism group of $GL_n(\mathbb{R})$ were $PGL_n(\mathbb{R})$ but I don't believe it is (for starters there will in general be an outer automorphism given by $X \mapsto (X^T)^{-1}$).