What i want to do is prove homotopy \begin{align} f \circ (\alpha * \beta) \simeq_{\{ f(x_0)\}} (f \circ \alpha) * ( f \circ \beta) \end{align} where $\circ$ is a composition $f \circ \alpha = f(\alpha)$ and $\alpha $ , $\beta$ is a loop based at $x_0$. $*$ is a path product, $\alpha * \beta(t) = \alpha(2t)$ for $0\leq t\leq \frac{1}{2}$, and $=\beta(2t-1)$ for $\frac{1}{2} \leq t \leq 1$.
I want to find $H(t,s)$ which satisfies
$H(0,s)= H(1,s) = f(x_0)$
$H(t,0) = f \circ (\alpha * \beta)$, $ H(t,1) = ( f \circ \alpha) * ( f \circ \beta)$.
Let $X$ and $Y$ be topological spaces, $x_0 \in X$, $f:X \to Y \ $ a continuous function, and $ \ \alpha : [0,1] \to X \ $ and $ \ \beta : [0,1] \to X \ $ (continuous) loops at $x_0$. Then $ \ \alpha(0) = \alpha(1) = x_0 \ $ and $ \ \beta(0) = \beta(1) = x_0$.
We have that $ \ \alpha * \beta : [0,1] \to X \ $ is a loop at $x_0$. In fact, it is continuous and $$(\alpha*\beta)(0) = \alpha(2 \cdot 0) = \alpha(0) = x_0 = \beta(1) = \beta(2-1) = \beta(2 \cdot 1 - 1) = (\alpha*\beta)(1) \ . $$ Of course $ \ f \circ \alpha : [0,1] \to Y$, $f \circ \beta : [0,1] \to Y \ $ and $ \ f \circ (\alpha * \beta) : [0,1] \to Y \ $ are continuous. Also we have that $$(f \circ \alpha)(0) = f \big( \alpha (0) \big) = f(x_0) = f \big( \alpha (1) \big) = (f \circ \alpha)(1)$$ and $$(f \circ \beta)(0) = f \big( \beta (0) \big) = f(x_0) = f \big( \beta (1) \big) = (f \circ \beta)(1) \ . $$ So, $f \circ \alpha$ and $f \circ \beta$ are loops at $f(x_0)$ and we are allowed to make the product $ \ (f \circ \alpha)*(f \circ \beta) : [0,1] \to Y$, which is a continuous loop at $f(x_0)$.
If $ \ 0 \leq t \leq 1/2$, then $$[f \circ (\alpha * \beta)](t) = f \big( (\alpha * \beta)(t) \big) = f \big( \alpha(2t) \big) = (f \circ \alpha)(2t) = [(f \circ \alpha) * (f \circ \beta)](t) \ . $$ If $ \ 1/2 \leq t \leq 1$, then $$[f \circ (\alpha * \beta)](t) = f \big( (\alpha * \beta)(t) \big) = f \big( \beta(2t-1) \big) = (f \circ \beta)(2t-1) = [(f \circ \alpha) * (f \circ \beta)](t) \ . $$ So, $[f \circ (\alpha * \beta)](t) = [(f \circ \alpha) * (f \circ \beta)](t)$, $\forall t \in [0,1]$. By definition, we have $$f \circ (\alpha * \beta) = (f \circ \alpha) * (f \circ \beta) \ . $$ Then, $f \circ (\alpha * \beta)$ is a continuous loop at $f(x_0)$.
We can choose $ \ H: [0,1] \times [0,1] \to Y \ $ such that, $\forall s,t \in [0,1]$, $$H(t,s) = [f \circ (\alpha * \beta)](t) = [(f \circ \alpha) * (f \circ \beta)](t) \ . $$ Then $H$ is continuous and, $\forall s \in [0,1]$, $$H(0,s) = [f \circ (\alpha * \beta)](0) = f(x_0) = [(f \circ \alpha) * (f \circ \beta)](1) = H(1,s) \ . $$ Also, $\forall t \in [0,1]$, $$H(t,0) = [f \circ (\alpha * \beta)](t) \ \ \ \ \text{ and } \ \ \ \ H(t,1) = [(f \circ \alpha) * (f \circ \beta)](t) \ . $$ Then $H$ is a homotopy from $f \circ (\alpha * \beta) \ $ to $ \ (f \circ \alpha) * (f \circ \beta)$ which fixes $f(x_0)$.