Short answer: the covariance matrix is a symmetric matrix, which is a hint that it is actually the matrix of a bilinear form.
Longer answer: any matrix is, by definition, the matrix of a linear transform - namely, the transform from $k^n$ to $k^n$ given by its coefficients! In a way, even a bilinear form is a transform matrix, from a space to its dual (the transform matrix $t$ associated to a bilinear form $b$ satisfies $\left< tx, y\right> = b(x,y)$, where $\left<,\right>$ is the canonical pairing between a space and its dual).
A better way to think of it is: what happens to this matrix when I apply a linear transform on its inputs ? For example, given a linear map $f: X \rightarrow Y$ and a linear transform $T$ on $X$, if $F$ is the matrix of $f$ and $F'$ the matrix of $F \circ T$, we have $F' = F \cdot T$ (as a matrix product). However, if $Q$ is a quadratic form on $X$, then it transforms as $Q' = \mathrm{transpose}(T) \cdot Q \cdot T$.
In the case of a covariance matrix, if we apply a transform $T$ on the inputs, we see that the covariance is modified in the same way as a bilinear form.