Transverse submanifolds have product charts

By [F2] choose a chart at $p$ for $S_1$ and let $f_1,\dots ,f_b$ be the last $b=m-\dim S_1$ coordinate functions; they are smooth, vanish on $S_1$ near $p$, and their differentials at $p$ are linearly independent and span the annihilator of $T_p{S}_1$. Similarly choose $g_1,\dots ,g_a$ from a slice chart of $S_2$, $a=m-\dim S_2$, spanning the annihilator of $T_p{S}_2$. Restricting to a common smaller neighbourhood of $p$, all these functions are defined there and still have the same properties at $p$.

Since $\dim S_1+\dim S_2=m$ and $T_p{S}_1+T_p{S}_2=T_{p}N$ by [F1], the sum is direct, $T_{p}N=T_p{S}_1\oplus T_p{S}_2$. Hence the annihilator of $T_p{S}_1$ and the annihilator of $T_p{S}_2$ meet only in $0$, and the $a+b=m$ covectors $d{g}_1(p),\dots ,d{g}_a(p),d{f}_1(p),\dots ,d{f}_b(p)$ are linearly independent: a linear relation splits into a combination of the $d{g}_j$ equal to minus a combination of the $d{f}_i$, which lies in the intersection of the two annihilators and hence vanishes, forcing all coefficients to vanish by independence in each family.

Within the slice chart of $S_1$ used in [F2], the functions $f_1,\dots ,f_b$ are $b$ of the coordinate functions, namely the last $b$, so their common zero locus inside that chart is exactly $S_1$ intersected with the chart domain; after restriction to the common smaller neighbourhood of step 1.1 this remains true there. The same holds for $g_1,\dots ,g_a$ and $S_2$.

Put $F:=(g_1,\dots ,g_a,f_1,\dots ,f_b)$ on the common neighbourhood of $p$, regarded as a smooth map into ${\mathbb{R}}^m$. By step 2.1 its differential at $p$ is an isomorphism, so by [F3] there is an open neighbourhood $U$ of $p$ such that $F{\mathrm{\mid }}_U$ is a diffeomorphism onto an open subset of ${\mathbb{R}}^m$; composing with a translation, $F{\mathrm{\mid }}_U$ is a chart $\varphi :(U,p)\to ({\mathbb{R}}^m,0)$ of the smooth manifold $N$ of [F4] at $p$.

In the chart $\varphi$ of step 3.1 the coordinates are the ordered functions $g_1,\dots ,g_a,f_1,\dots ,f_b$; by step 2.2, $\varphi (S_2\cap U)=\{x_1=\cdots =x_a=0\}\cap \varphi (U)$ and $\varphi (S_1\cap U)=\{x_{a+1}=\cdots =x_m=0\}\cap \varphi (U)$. Writing $a=\dim S_1$ and $b=\dim S_2$ and identifying the last $b$ coordinates with ${\mathbb{R}}^b$ and the first $a$ with ${\mathbb{R}}^a$ gives exactly $\varphi (S_1\cap U)={\mathbb{R}}^a\times \{0\}\cap \varphi (U)$ and $\varphi (S_2\cap U)=\{0\}\times {\mathbb{R}}^b\cap \varphi (U)$.

For the general clause use the sum map instead of defining functions. Identify a neighbourhood of $p$ with ${\mathbb{R}}^m$ by a chart [F4] carrying $p$ to $0$, and consider $H:S_1\times \cdots \times S_r\to {\mathbb{R}}^m$, $H(s_1,\dots ,s_r)=s_1+\cdots +s_r$. Its differential at $(p,\dots ,p)$ sends $(v_1,\dots ,v_r)$ to $v_1+\cdots +v_r$, which is invertible exactly because $T_{p}N=T_p{S}_1\oplus \cdots \oplus T_p{S}_r$; by [F3] $H$ restricts to a diffeomorphism from a neighbourhood of $(p,\dots ,p)$ onto an open neighbourhood $U$ of $p$, with smooth inverse $H^{-1}(q)=(a_1(q),\dots ,a_r(q))$, $a_j(q)\in S_j$. Shrink $U$ so that for each $i$, every $q\in S_i\cap U$ lies in the chosen factor neighbourhood and $(p,\dots ,q,\dots ,p)$ lies in the inverse-function domain. Since $H$ is injective there and $H(p,\dots ,q,\dots ,p)=q$, a point $q\in U$ lies in $S_i$ exactly when $a_j(q)=p$ for every $j\mathrm{\ne }i$: if $q\in S_i$ then $(p,\dots ,q,\dots ,p)$ and $(a_1(q),\dots ,a_r(q))$ are two preimages of $q$, hence equal, and conversely $a_j(q)=p$ for $j\mathrm{\ne }i$ gives $q=a_i(q)\in S_i$. Composing $H^{-1}$ with charts of the $S_j$ at $p$ given by [F2] produces a chart $\varphi =({\varphi }^{(1)},\dots ,{\varphi }^{(r)})$ of $N$ at $p$, ${\varphi }^{(j)}$ taking values in ${\mathbb{R}}^{\dim S_j}$, and the criterion just proved says $\varphi (S_i\cap U)=\{{\varphi }^{(j)}=0,j\mathrm{\ne }i\}\cap \varphi (U)$, a coordinate subspace.