Normal Bundle over Transverse Intersection From Bott and Tu's Differential Forms in Algebraic Topology:

Two submanifolds $R$ and $S$ in $M$ are said to intersect transversally iff $$T_xR + T_x S = T_x M$$ for all $x \in R \cap S$. For such a transversal intersection the codimension in $M$ is additive: $$\text{codim } R \cap S = \text{codim } R + \text{codim } S$$
  This implies that the normal bundle of $R \cap S$ in $M$ is $$N_{R \cap S} = N_R \oplus N_S$$

I don't understand how this implication works. I understand it intuitively, but cannot see how it follows from the codimension thing. Any help would be appreciated.
 A: I'll assume that normal bundles are defined as quotient objects, so for example $(N_R)_x = (T_x M) / (T_x R)$. From this it is clear that the injections $T_x(R \cap S) \hookrightarrow T_x R \hookrightarrow T_x M$ induce an injection $(N_R)_x \hookrightarrow (N_{R \cap S})_x$. Similarly, $(N_S)_x \hookrightarrow (N_{R \cap S})_x$. So, we may abuse notation and identify $(N_R)_x$ and $(N_S)_x$ with their embedded images in $(N_{R \cap S})_x$.
Furthermore, using the equation $T_x M = T_x R + T_x S$, together with a little algebra, we can deduce that
$$(*) \qquad (N_{R \cap S})_x = (N_R)_x + (N_S)_x
$$
(and it is here that I am using the notational abuse).
Next, we have $\text{codim} R \cap S = \text{dim} N_{R \cap S}$; and similarly for $R$ and for $S$. It follows that
$$(**) \quad \dim N_{R \cap S} = \dim N_R + \dim N_S
$$
Combining $(*)$ and $(**)$ we get
$$(N_{R \cap S})_x = (N_R)_x \oplus (N_S)_x
$$
Added: Here are some lemmas from linear algebra (the theory of vector spaces) which I used in deducing (*). You should be able to find these in any advanced linear algebra textbook.

If $V$ is a vector space, if $A,B \subset V$ be two subspaces, and if $q : V \to V /(A \cap B)$ is the natural quotient map, and if $V = A + B$, then the following hold:

*

*$q|A$ induces an isomorphism from $A / (A \cap B)$ to $q(A)$.

*$q|B$ induces an isomorphism from $B / (A \cap B)$ to $q(B)$

*$V / (A \cap B) = q(A) + q(B)$.

Therefore, using $q$ to identify $A / (A \cap B) \approx q(A)$ and $B / A \cap B \approx q(B)$, we obtain
$$V / (A \cap B) = A / (A \cap B) + B / (A \cap B)
$$

