Homotopicity of two certain sections of frame bundle of $GL(n,\mathbb{R})$ Edit: According to comment of Prof. GoodWillie we revise the question.
Put $M=GL(n,\mathbb{R})$.
We identify $M_n(\mathbb{R})$ with $\mathbb{R}^{n^2}$:
The identification is based on the lexicographic order on the index $i,j$ in $(a_{ij})$. For example $$ \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$$
is identified with $$(a_{11}, a_{12},a_{21}, a_{22})$$
So $M$ being an open subset of $\mathbb{R}^{n^2}$ has trivial tangent bundle and there is an obvious description for a tangent vector at a point $A\in M$.
The mapping $A\mapsto A\otimes A$ defines a section of fram bundle of $M$.Because each row of $A\otimes A$ is counted as a $n×n$ matrix via the above identification. So $A\otimes A$ actualy determines $n^2$ independent vectors in the tangent space to $M$ at $A$.
In a similar manner $A\otimes A^{tr}$ is another section of the frame bundle of manifold $M$ where $tr$ is transpose operator.

Are the above two sections $A\otimes A$ and $A\otimes A^{tr}$ homotopic sections of frame bundle of $M$?

 A: They are homotopic when $n=2$, but not when $n>2$.  Here is the argument:
Let $A=(a_{ij})$.  Then, the definitions of the two framings can be made more explicit as follows:  For the first frame field, each vector field, say, $X_{ij}$, can be thought of as an $n$-by-$n$ matrix, and the formula for the $kl$ entry of $X_{ij}$ is
$$
(X_{ij})_{kl} = a_{ik}a_{jl}\,,
$$
while for the second frame field, each vector field, say $Y_{ij}$, can be thought of as an $n$-by-$n$ matrix, and the formula for the $kl$ entry of $Y_{ij}$ is
$$
(Y_{ij})_{kl} = a_{ik}a_{lj}\,.
$$
The $1$-forms $\xi^{ij}$ of the coframing dual to the framing $X_{ij}$ are then seen to be the components of the $n$-by-$n$ matrix-valued $1$-form
$$
\xi = ({}^T\!\!A)^{-1} \mathrm{d}A\, A^{-1} = (\xi^{ij})
$$ 
while the $1$-forms $\eta^{ij}$ of the coframing dual to the framing $Y_{ij}$ are then seen to be the components of the $n$-by-$n$ matrix-valued $1$-form
$$
\eta = ({}^T\!\!A)^{-1} \mathrm{d}A\, ({}^T\!\!A)^{-1} = (\eta^{ij}).
$$
Now, the framings $X$ and $Y$ are homotopic if and only if the coframings $\xi$ and $\eta$ are homotopic.  Since $\eta = \xi A ({}^T\!\!A)^{-1}$, it follows that these framings are homotopic over $\mathrm{GL}(n,\mathbb{R})$ if and only if the map $\phi:GL(n,\mathbb{R})\to \mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))\simeq \mathrm{GL}(n^2,\mathbb{R})$ defined by
$$
\phi(A) = R\bigl(A({}^T\!\!A)^{-1}\bigr)
$$
is null-homotopic, where $R(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$ is right multiplication by $B$, i.e., $R(B)C = CB$.  Note that $R(B)$ is simply $n$ copies of the natural right action of $B$ on $\mathbb{R}^n$, so $\phi$ is actually equivalent to the mapping
$$
A\mapsto \bigl(A({}^T\!\!A)^{-1},\ A({}^T\!\!A)^{-1}, A({}^T\!\!A)^{-1}\,,\ldots,\  A({}^T\!\!A)^{-1}\bigr)
$$
i.e., $n$ diagonal copies of the mapping $\psi:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ defined by $\psi(A) =  A({}^T\!\!A)^{-1}$.
Now, as is well known, $\mathrm{GL}(n,\mathbb{R})$ is diffeomorphic to $S^+_n\times \mathrm{O}(n)$, where $S^+_n$, which is contractible, is the space of $n$-by-$n$ positive definite matrices.  This is the famous $QR$-decomposition, i.e., $A = QR$ where $Q$ is symmetric positive definite and $R$ is orthogonal.  Since $S^+_n$ is contractible, we can answer the homotopy question by restricting $\phi$ to $O(n)$, i.e., we set $Q=I_n$.  On $\mathrm{O}(n)$ the mapping becomes
$$
\phi(R) = (R^2,\ R^2,\ ,\ldots,\ R^2)\in \mathrm{SO}(n^2)
$$
Note that the image goes diagonally into $\mathrm{SO}(n)\times\cdots\times\mathrm{SO}(n)\subset \mathrm{SO}(n^2)$.
Now, when $n=2$, $\mathrm{SO}(2)\simeq S^1$, and $\pi_1\bigl(\mathrm{SO}(2)\bigr)\simeq\mathbb{Z}$.  Meanwhile, $\pi_1\bigl(\mathrm{SO}(4)\bigr)\simeq\mathbb{Z}_2$, and $\phi$ takes a generator of $\pi_1\bigl(\mathrm{SO}(2)\bigr)$ to $4$ times a generator of $\pi_1\bigl(\mathrm{SO}(4)\bigr)$, i.e., to zero.  Thus, when $n=2$, the mapping $\phi$ is null-homotopic, so the two framings $X$ and $Y$ are homotopic when $n=2$.
However, when $n>2$, we have $\pi_3\bigl(\mathrm{SO}(n)\bigr)\simeq\mathbb{Z}$ (except when $n=4$, when it equals $\mathbb{Z}\oplus\mathbb{Z}$).  Moreover, it is easy to see that, when $n\not=3$, the mapping $\phi$ takes a generator of $\pi_3\bigl(\mathrm{SO}(n)\bigr)$ to $2n$ times a generator of $\pi_3\bigl(\mathrm{SO}(n^2)\bigr)$, which is nontrivial.  Thus, $\phi$ is not homotopically trivial in these cases. A similar argument applies in the case $n=4$, to show that $\phi$ is not homotopically trivial in this case either.  Thus, when $n>2$, the two framings are not homotopic.
A: The two sections are not homotopic for large $n$. Here's why.
First, as you note, the tangent bundle of $GL_{n}$ is trivial, which means its frame bundle is as well. Next, as noted in the comments, we might as well replace $GL_{n}$ with $O(n)$, which is nicer because it is easier to typeset, and we can swap the transpose operator for the inverse operator.
The two sections you define are therefore maps $s,t:O(n)\rightarrow O(n)\times O(n^{2})$ given by $A\mapsto (A, A\otimes A)$ and $A\mapsto (A, A\otimes A^{-1})$. So to check homotopicity it's enough to check homotopicity of the maps $O(n)\rightarrow O(n^{2})$ that you get from the sections after projecting on to the second factor.
Let's check what the maps do on $\pi_{3}$. First note that $\pi_{3}O(n)=\mathbb{Z}$ for all $n>5$. Next, let's factor our maps $s=S\circ\Delta$, $t=T\circ\Delta$ where $\Delta:O(n)\rightarrow O(n)\times O(n)$ is the diagonal, $S(A,B)=A\otimes B$, and $T(A,B)=A\otimes B^{-1}$.
As Tom Goodwillie noted in the comments, the inverse map acts as $-1$ on all homotopy groups.
So the maps induced by $S$ and $T$ are two $1\times2$ integer matrices, and the second column of $T$ is the negative of the second column of $S$; i.e. $S=[a\ \ b]$ and $T=[a\ \ -b]$. If $s=t$ (on homotopy) then $a+b=a-b$ so $S=[a\ \ 0]$. But that clearly can't be, for symmetry reasons.
