Compatible Hilbert space subspaces - need help understanding a statement made in a book

No, the book doesn't have any typo.

First let me give you an example to have an imagination of what is going on. The simplest example to imagine is the following. Let $M_1$ be the $x$-axis in $\mathbb{R}^3$, $N_1$ be the $y$-axis, $K$ be the $z$-axis. Obviously $M_1$, $N_1$ and $K$ are mutually disjoint subspaces of $\mathbb{R}^3$, now you can see that $M$ is the $x$o$z$ plane and $N$ is the $y$o$z$ plane. And all those three relations $K=M\cap N$, $M_1=M\cap K^{\perp}$ and $N_1=N\cap K^{\perp}$.

But now for proving the two last that you have doubt about them. I will do for the $M_1$ and one for the $N_1$ will follow with same story.

First consider that because $M_1$ and $K$ are mutually disjoint, and also as $M=M_1+K$; $$\left.\begin{array}{l} M_1\subseteq M\\ M_1\subseteq K^{\perp}\end{array}\right\}\Longrightarrow M^{\perp}\subseteq M\cap K^{\perp}$$

Now for the opposite inclusion, chose and arbitrary element in $M\cap K^{\perp}$, say $v$. Since $v\in M=M_1+K$ we can right it as $v=a+b$ where $a\in M_1$ and $b\in K$. Since $v\in K^{\perp}$ we have for every vector in $K$, say $k$, the inner product of $v$ and $k$ should be zero. $$0=\langle a+b,k\rangle=\langle a,k\rangle+\langle b,k\rangle=0+\langle b,k\rangle$$ Inner product of $a$ and $k$ is zero because $a$ is in $M_1$ and $M_1$ is subset of orthogonal completion of $K$. And no element of a space except the trivial vector $0$ is normal to that space so for the equation $\langle b,k\rangle=0$ be hold, we have to have $b=0$ and this shows $v=a\in M_1$. So we showed $M\cap K^{\perp}\subseteq M_1$.