The condition for $E$ to be intransitive is that the determinant form is the $0$ form somewhere other than the origin. That is, every vector $v in mathbb R^d$ gives you an alternating $d$-form on $M_d$ and on $E$ by the determinant of the images of $v$. This form is nonzero on a vector if and only if the images of the vector by $E$ are all of $mathbb R^d$.
Edit: You are right that the above was not a complete answer.
To be more explicit with bases for everything: Let $E$ have dimension $D$, with $d le D le d^2$. Let $E$ have a basis ${E_1, ... E_D }$ so that every element of $E$ is represented by a vector $(a_1,...,a_D)$. Represent every element of $mathbb R^d$ by a vector $(b_1,....,b_d)$.
Then for any vector $(b_1,....,b_d)$, the determinant form is an alternating $d$-form on $E$ identified with $mathbb R^D$. These forms have a basis of size ${D choose d}$ given by the determinants of $dtimes d$ minors, that is, project to a given $d$ coordinates, and take the determinant. To check whether the determinant form is the $0$ $d$-form, express it in terms of the basis, and see if all ${D choose d}$ coefficients are $0$. That is, check if the ${D choose d}$ determinants $det [E_{f(1)}v, ..., E_{f(d)}v]$ are all $0$ for each integer-valued function $f$ with
$1 le f(1) lt f(2) lt ... lt f(d) le D$.
As we let $v$ vary but fix a basis for $E$, the coefficients of the determinant form are homogeneous polynomials of degree $d$ in the coordinates ${b_i}$. The variety of intersections of the zeros of those polynomials on $mathbb R^d$ is the origin if and only if $E$ is transitive.
This gives you a test for a particular $E$ in terms of recognizing whether a variety is just a point. It still leaves the condition on $E$ in the Grassmannian as a projection of a variety.
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