One way to describe the embedding $Spin_9\to F_4$ is as follows. $F_4$ is the group of automorphisms of the Jordan algebra $\mathcal J$ of traceless (or, equivalently, matrices with any given constant (real) trace, since an automorphism must fix the the real multiples of the identity and preserve the trace) Hermitian $3\times 3$ matrices with Cayley numbers (octonions) as coefficients. The algebra of Cayley numbers is a non-associative normed division algebra over $\mathbf R$ of dimension $8$. It follows that $\mathcal J$ is a real vector space of dimension $3\cdot 8 + 2 = 26$ ($3-1=2$ for the diagonal entries, which are real and add up to zero). This gives a $26$-dimensional representation of $F_4$ which is irreducible ($\omega_4$ is the notation of Bourbaki) and in fact the lowest dimensional non-trivial representation of $F_4$. The structure of Jordan algebra is given by $a\cdot b = \frac12(ab+ba)$ for $a$, $b\in\mathcal J$ (on the right hand side we have the usual multiplication of matrices, but recall that the Cayley algebra is neither associative, nor commutative). Now let $$p=\left(\begin{array}{ccc}1&0&0;\\ 0&0&0; \\ 0&0&0\end{array}\right)\in\mathcal J.$$ Then the isotropy subgroup of $F_4$ at $p$ is isomorphic to $Spin_9$. The orbit through $p$ is thus $F_4/Spin_9$, namely, the Cayley projective plane.

The representation $\mathcal J$ is in some sense what you need to get $E_6$ from $F_4$. In fact, $\mathfrak e_6=\mathfrak f_4 + \mathcal J$ as vector spaces, where the Lie algebra structure is as follows: for $X$, $Y\in \mathfrak f_4$, $a$, $b\in \mathcal J$ we have that $[X,Y]$ is the bracket in $\mathfrak f_4$, $[X,a]=Xa$ is the action $\rho$ of  $\mathfrak f_4$ on $\mathcal J$ and $[a,b]\in\mathfrak f_4$ is defined by $\langle X,[a,b]\rangle = \langle Xa,b\rangle$ for all $X\in \mathfrak f_4$ and $\langle X,Y\rangle = Kill_{\mathfrak f_4}(X,Y) + trace(\rho(X)\rho(Y))$. This is in fact a strandard construction, taking into account that $F_4$ is a symmetric subgroup of $E_6$ and $E_6/F_4$ is a symmetric space.

There are other things but at this point I'd refer you to the very nice book "Lectures on Exceptional Lie Groups", J. F. Adams Edited by Zafer Mahmud and Mamoru Mimura, 1996, Chicago Lectures in Mathematics.

One way to describe the embedding $Spin_9\to F_4$ is as follows. $F_4$ is the group of automorphisms of the Jordan algebra $\mathcal J$ of traceless (or, equivalently, matrices with any given constant (real) trace, since an automorphism must fix the the real multiples of the identity and preserve the trace) Hermitian $3\times 3$ matrices with Cayley numbers (octonions) as coefficients. The algebra of Cayley numbers is a non-associative normed division algebra over $\mathbf R$ of dimension $8$. It follows that $\mathcal J$ is a real vector space of dimension $3\cdot 8 + 2 = 26$ ($3-1=2$ for the diagonal entries, which are real and add up to zero). This gives a $26$-dimensional representation $\rho$ of $F_4$ which is irreducible ($\omega_4$ is the notation of Bourbaki) and in fact the lowest dimensional non-trivial representation of $F_4$. The structure of Jordan algebra is given by $a\cdot b = \frac12(ab+ba)$ for $a$, $b\in\mathcal J$ (on the right hand side we have the usual multiplication of matrices, but recall that the Cayley algebra is neither associative, nor commutative). Now let $$p=\left(\begin{array}{ccc}1&0&0;\\ 0&0&0; \\ 0&0&0\end{array}\right)\in\mathcal J.$$ Then the isotropy subgroup of $F_4$ at $p$ is isomorphic to $Spin_9$. The orbit through $p$ is thus $F_4/Spin_9$, namely, the Cayley projective plane.

The representation $\mathcal J$ is in some sense what you need to get $E_6$ from $F_4$. In fact, $\mathfrak e_6=\mathfrak f_4 + \mathcal J$ as vector spaces, where the Lie algebra structure is as follows: for $X$, $Y\in \mathfrak f_4$, $a$, $b\in \mathcal J$ we have that $[X,Y]$ is the bracket in $\mathfrak f_4$, $[X,a]=\rho(X)a$ is the action $\rho$ of $\mathfrak f_4$ on $\mathcal J$ and $[a,b]\in\mathfrak f_4$ is defined by $\langle X,[a,b]\rangle = \langle \rho(X)a,b\rangle$ for all $X\in \mathfrak f_4$ and $\langle X,Y\rangle = Kill_{\mathfrak f_4}(X,Y) + trace(\rho(X)\rho(Y))$. This is in fact a standard construction, taking into account that $F_4$ is a symmetric subgroup of $E_6$ and $E_6/F_4$ is a symmetric space.

There are other things but at this point I'd refer you to the very nice book "Lectures on Exceptional Lie Groups", J. F. Adams Edited by Zafer Mahmud and Mamoru Mimura, 1996, Chicago Lectures in Mathematics.