Let $F(f)$ be the homotopy fiber of the map $X\to Y$. Then there is a fiber sequence $F(f)\xrightarrow{p}X\xrightarrow{f}Y$, where $p:(x,\gamma)\mapsto x$. I wonder why the map $fp$ is (pointed) null homotopic? $fp(F(f))=f(X)\subset Y$, $\pi_1(f(X))$ is not guaranteed to be trivial, why is the $fp$ null-homotopic?
Reference:http://www.home.uni-osnabrueck.de/mfrankland/Math527/Math527_0304.pdf Proposition 2.1
In the notes you linked, $F(f)$ is defined as the pullback of the diagram $X \xrightarrow{f} Y \leftarrow PY$ where $PY$ is the path-space of $Y$. Thus the map $fp$ is the same as the map $F(f) \to PY \to Y$. But $PY$ is contractible, so the map is nullhomotopic.
To see that $PY$ is contractible, note that we can define for a path $\gamma(s)$ the associated path $\gamma_t(s) = \gamma(ts)$ (shortening the path to a fraction $t$ of its total length). You can check that the map $\gamma \mapsto \gamma_t$ is a deformation retraction from $PY$ to the constant path at the basepoint $y_0$.