$\newcommand{\Ind}{\operatorname{Ind}} \newcommand{\Gr}{\operatorname{Gr}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\R}{\mathbb{R}} \newcommand{\GL}{\operatorname{GL}} \newcommand{\codim}{\operatorname{codim}}$ Let $\Gr_k$ be the Grassmannian manifold of $k$-dimensional subspaces in $\R^n$. Let $\GL(n)$ act on $\Gr_k$ in the natural way. For an arbitrary such subspace $V \in \Gr_k$, let $\varphi_V \colon \GL(n) \to \Gr_k$ take $\varphi_V(g) = g V$. This is a surjective map with surjective differential.
Differentiating $\varphi_V$ at the identity element $e \in \GL(n)$ gives a map $d \varphi_V \colon \Hom(\R^n, \R^n) \to T_V \Gr_k$, and its kernel are the maps $f$ that stabilize $V$, in the sense that $f(V) \subseteq V$. Since the map $\eta_V \colon \Hom(\R^n, \R^n) \to \Hom(V, \R^n / V)$ defined by $\eta_V(f) = \pi_V \circ f \circ \iota_V$ has the same kernel, this establishes an isomorphism $T_V \Gr_k \cong \Hom(V, \R^n / V)$.
When $W$ is some subspace, define $\Ind_k(W)$ to be the subset of elements $V \in \Gr_k$ that intersect $W$ in dimension at least one. In the literature, this is a special case of a "Schubert variety." I strongly suspect that $\Ind_k(W)$ is an embedded submanifold of $\Gr_k$ in a neighborhood of an element $V$ where $\dim V \cap W = 1$. Furthermore, I believe its tangent space at such an element is composed, under the isomorphism above, of the maps $f$ for which $f(V \cap W) \subseteq W / V$.
However, I'm having trouble proving this rigorously. One idea is to consider the function $F(g) = g V \wedge W$, viewing $V$ and $W$ as arbitrarily scaled elements of the exterior algebra $\Lambda(\R^n)$. The system of $k$-degree polynomial relations $F(g) = 0$ cuts out the preimage of $\Ind_k(W)$ under $\varphi_V$. We can also differentiate $F$ at the identity in the following way. Since $V \cap W$ is a one-dimensional subspace, we have a basis $v_1, \ldots, v_k$ for $V$ with $V \cap W = \langle v_1 \rangle$. For $A \in \Hom(\R^n, \R^n)$, differentiating gives \begin{align*} d F_e(A) & = \frac d{dt}_{t = 0} e^{t A} V \wedge W = \frac d{dt}_{t = 0} \left (\bigwedge_{i = 1}^k e^{t A} v_i \right) \wedge W \\ & = \sum_{i = 1}^k (-1)^{i + 1} A v_i \wedge \bigwedge_{\substack{j = 1 \\ j \neq i}}^k v_j \wedge W = A v_1 \wedge v_2 \wedge \ldots \wedge v_k \wedge W, \end{align*} where the last equality follows because $v_1 \wedge W = 0$. In particular, we conclude that the kernel of $dF$ are the maps $f$ for which $f(V \cap W) \subseteq V + W$. The image of $\ker dF_e$ under $d \varphi_V$ coincides with the description of $T_V \Ind_k(W)$ that I have conjectured.
Unfortunately, I believe that $F$ does not have constant rank near the identity. How can I easily show that $F^{-1}(0)$ is an embedded submanifold near this point, with tangent space given by the kernel of $dF$?
(The following is my best answer at the moment, but any simpler / different approach is also eligible for the bounty. I'd also like to know if there is a more sophisticated justification from algebraic geometry for this kind of thing to happen—namely, for a variety defined by a system of polynomials not respecting the CRT to nevertheless have smooth points almost everywhere, with tangent spaces corresponding to the kernel of the system's differential.)
$\newcommand{\Ind}{\operatorname{Ind}}\newcommand{\codim}{\operatorname{codim}}\newcommand{\GL}{\operatorname{GL}}\newcommand{\R}{\mathbb{R}}\newcommand{\rank}{\operatorname{rank}}$In fact, $F$ does not have constant rank near the identity. Fortunately, the computation of the tangent space to $\Ind_k(W)$ using $F$ can be salvaged. For this, we will show that there is a linear map $\pi$ of rank $\codim V + W$ so that the equation $(\pi \circ F)(g) = 0$ has the same solutions for $g$ in a neighborhood of $e$, and so that $\pi \circ dF_e$ is surjective. Then the constant rank theorem can be applied to $\pi \circ F$ will show that $F^{-1}(0)$ is an embedded submanifold of $\GL(n)$.
The essential idea can be described in terms of matrices. Consider the determinantal variety $D$ of $m \times n$ matrices, $m \le n$, with rank $m - 1$. A matrix $M$ is an element of $D$ exactly when all its $\binom{n}{m}$ submatrices of size $m \times m$ are singular. However, in a neighborhood of any $M \in D$, only $n - m + 1$ determinants suffice to define $D$. For example, suppose that the first $m - 1$ columns of $M$ are linearly independent. Then the same is true of any matrix $N$ in a neighborhood of $M$. Furthermore, if $N$ has rank $m$, then a basis for its column space can be formed with its first $m - 1$ columns and one of its last $n - m + 1$ columns. In other words, $N$ has rank $m - 1$ exactly when all $n - m + 1$ square $m \times m$ submatrices containing its first $m - 1$ columns are singular. In fact, the differentials of these polynomials will be linearly independent in a neighborhood of $M$.
We can also state this in exterior algebra language. Let $v_1, \ldots, v_k$ be a basis for $V$, $v_1, w_2, \ldots, w_d$ be a basis for $W$, and $x_1, \ldots, x_\ell$ be a basis for a complementary subspace to $V + W$, so that $$\{ v_1, \ldots, v_k, w_2, \ldots, w_d, x_1, \ldots, x_\ell \}$$ form a basis for $\R^n$. Consider the multivector basis for $\Lambda^{k + d}(\R^n)$ corresponding to this basis for $\R^n$. For each $i = 1, \ldots, \ell$, let $B_i$ the vector $x_i$ appended to our basis for $V + W$. We claim that the projection $\pi$ of $\Lambda^{k + d}$ onto the basis multivectors $\bigwedge_{b \in B_i} b$ has the properties we have promised.
Indeed, for $g$ in a neighborhood of $e$, we know the projection of $g V + W$ under our basis for $V + W$ is full-dimensional. We will have $\dim g V \cap W = 0$ exactly when $\dim g V + W = k + d$, in which case the coordinate projection of $g V + W$ under some basis $B_i$ will be full-dimensional. This is the same as $\bigwedge_{b \in B_i} b$ appearing with non-zero coefficient in the expansion of $g V \wedge W$ under our basis. (In fancypants language, the projection onto $\bigwedge_{b \in B_i} b$ can be viewed as the $(k + d)$th exterior power of the projection onto $B_i$, and thus is a functional on multivectors telling us whether their corresponding subspaces project surjectively onto $B_i$.) Thus, in some neighborhood of $e$, $\varphi_V(g) \in \Ind_k(W)$ is equivalent to $\pi(F(g)) = 0$. Given our expression for $d F_e$, it is easy to check that $\rank \pi \circ d F_e = \ell = \rank \pi$ as desired.