Another point of view. One can construct very explicitely $f_{4}$ by beginning with
$so(9)$. Let $J_{ij}$ be a basis of $so(9)$. Consider the 16-dimensional spin representation of
$so(9)$ with a basis $Q^{a}$ : one has $[J_{ij},Q^{a}]= \gamma_{ab,ij}Q_{b}$. Then consider the direct sum of $so(9)$ and of its spin representation. We want to construct a Lie algebra structure on this vector space. For $[J_{ij}, J_{kl}]$ one takes what we have in $so(9)$,
and one set $[J_{ij},Q^{a}]= \gamma_{ab, ij}Q_{b}$ and
$[Q^{a}, Q^{b}] = \gamma_{ab, ij} J_{ij}$.

There is some computation to do in order to prove that Jacobi identity is verified.
The Lie algebra obtained is of dimension 36+16=52 and is $f_{4}$.
By construction, $so(9)$ is a Lie subalgebra of $f_{4}$.

The same construction with $so(16)$ and its 128-dimensional spin representation gives a Lie algebra of dimension 120+128=248 which is $e_{8}$. The natural inclusion of $so(9)$ in $so(16)$
induces an inclusion of $f_{4}$ in $e_{8}$. In fact $e_{8}$ contains naturally $e_{6}$ : $e_{6}$ is the centralisator of $so(10) \times su(3) \subset so(10) \times su(4)=so(10) \times so(6) \subset so(16) \subset e_{8}$.
With this explicit description, it is easy to see that $f_{4}$ is in fact contained in $e_{6}$.