Here's a motivating question: suppose $N_1, N_2$ are transversely intersecting submanifolds of $M$ and let $p$ be a point in the intersection. Does there exist a chart $(U, \varphi)$ of $M$ at $p$ such that the diffeomorphism $\varphi : U \to \Bbb R^n$ takes $N_1 \cap U$ and $N_2\cap U$ to two transverse subspaces of $\Bbb R^n$, i.e., $\varphi(N_i \cap U) = H_i$ where $H_i \subset \Bbb R^n$ are transverse subspaces of $\Bbb R^n$ with $\dim H_i = \dim N_i$?
This is more or less asking if, similar to the immersion and submersion theorem which are essential to differential topology, transversality also have a local model based on the picture $(\Bbb R^n, H_1, H_2)$ consisting of two transverse subspaces $H_1,H_2 \subset \Bbb R^n$.
There are probably easier proofs of this fact, but my sketch was as follows. Suppose we impose a Riemannian metric $g$ on $M$ so that $N_1$ and $N_2$ becomes totally geodesic. Consider the exponential map $\exp_p : T_p M \dashrightarrow M$. Since $N_1$ and $N_2$ are transverse submanifolds, there's a decomposition $T_pM = T_pN_1 + T_pN_2$, and $\exp_p$ restricts to the two factors of this decomposition (for any $v \in T_p N_i$ in the domain of definition, $\exp_p(tv)$ is the geodesic starting at $p$ with initial vector $v$. Under the induced Riemannian metric there is also a geodesic $\gamma$ of $N_i$ starting at $p$ with initial vector $v$ - since $N_i$ is totally geodesic $\gamma$ is also a geodesic of $M$, so by uniqueness we have to have $\gamma(t) = \exp_p(tv)$, forcing $\exp_p(v)$ to map to $N_i$.) Since $\exp_p$ is a local diffeomorphism at the origin, one can choose a ball $B$ around $0$ of radius $\varepsilon$ in $T_pM$ so that $\exp_p : B \to M$ is a diffeomorphism which sends the transverse subspaces $H_i := T_p N_i \cap B$ diffeomorphically to $N_i$, as desired
This brings me to the question of the title: Given a finite family $N_\alpha$ of mutually transversely intersecting submanifolds of $M$, when is it possible to impose a Riemannian metric $g$ on $M$ such that $N_\alpha$ are totally geodesic submanifolds of $(M, g)$?
For a single submanifold $N \subset M$, this should be doable as follows: take the normal bundle $\pi: E \to N$. Choose a metric $g_N$ on the base and a bundle metric $\tilde{g}$ for $\pi$. Moreover choose a connection on $\pi$ compatible with the bundle metric so that there's a canonical decomposition $TE \cong H \oplus V$ into horizontal and vertical subbundles. Define the bundle metric on $d\pi : TE \to TM$ as $g_E((x_h, x_v), (y_h, y_v)) = g_N(x_h, y_h) + \tilde{g}(x_v, y_v)$ where the subscripts denote the horizontal($h$) or vertical($v$) coordinates respectively.
This makes the total space into a Riemannian manifold $(E, g_E)$ and indeed one so that the map $\iota : E \to E$ defined by $\iota((x, v)) = f(x) \cdot v$ for a chosen scalar function $f : N \to\Bbb R$ is an isometry. This is nothing but scaling fiberwise by a constant, which leaves the zero section $E_0 \subset E$ fixed. Fixed point loci of isometries of a Riemannian manifold are totally geodesic submanifold, so this guarantees $E_0 \subset E$ is totally geodesic.
But by the tubular neighborhood theorem, there is a neighborhood $U$ of $N \subset M$ such that $U$ is diffeomorphic to a neighborhood of the zero section $E_0 \subset E$. Therefore $U$ admits a metric $g_U$ obtained from pulling back $g_E$ in which $N$ is a totally geodesic submanifold of $(U, g_U)$. Let $V := M \setminus N$ and give $V$ an arbitrary Riemannian metric. Use partition of unity on the cover $\{U, V\}$ of $M$ to glue these two metrics to a metric $g$ on $M$ such that $N \subset (M, g)$ is totally geodesic.
One immediately runs into trouble trying to generalize this argument for multiple transverse submanifolds of $M$ as the glued metrics would change near the points of intersection, no longer assuring total geodesicity. Is this true in general? If so, what would be a proof?