A set of notes I am following is defining the curvature of a connection. It does so by first extending the connection operator $D: \Gamma(E) \to \Omega^1 E$, where $(E,M,\pi)$ is a vector bundle, to a more general operator between the spaces $\Omega^p E \to \Omega^{p+1}E$, here $\Omega^p(E):=\Gamma(E\otimes\Lambda^pM)$. The notes extend $D$ by imposing the Leibniz rule;
$D(\omega \otimes\sigma)=d\omega \otimes \sigma+(-1)^p\omega\wedge D\sigma$
$\omega \in \Omega^pM $, $\sigma \in \Gamma(E)$.
In the notes the above Leibniz rule is then use to show the tonsorial nature of the operator $D^2$ which is later defined as the curvature of the connection. i.e. the notes show that $D^2(fs)=fD^2(s)$, $\forall s \in\Gamma(E), f \in C^\infty(M)$
I have tried this but unfortunately do not agree with their calculation. I tried;
\begin{align*} D^2(fs)&=D(df \otimes s+fDs) \hspace{0.2in}\text{(Leibniz)}\\ &=D(df \otimes s)+D(fDs) \hspace{0.2in}\text{(Linearity)}\\ &=d^2f \otimes s-df \wedge Ds+df \otimes Ds+fD^2s \\ \end{align*} Where as my notes say that I should get $$D^2(fs)=d^2f \otimes s-df \otimes Ds+df \otimes Ds+fD^2s $$ which would complete the proof.
I don't see any reason why $df \wedge Ds$ should be equal to $df \otimes Ds$. Is the Leibniz rule provided in my notes wrong? Or in this case is $df \wedge Ds$ equal to $df \otimes Ds$?
Here's your mistake: the formula for $D(\omega\otimes\sigma)$ that you're using applies only when $\sigma$ is a section of $E$ (i.e., an $E$-valued $0$-form). But when you expand $D(fDs)$, the factor $Ds$ is not a section of $E$, it's an element of $\Omega^1E$ (an $E$-valued $1$-form).
You should try first to use the definition to prove the following formula for all $\omega\in \Omega^pM$ and $\sigma\in \Omega^qE$: $$ D(\omega\wedge\sigma) = d\omega \wedge \sigma + (-1)^p \omega \wedge D\sigma. $$ (To compute the wedge product of $\omega$ with a section of $\Omega^{q}E$, just write the latter locally as a sum of terms like $\omega_i \otimes \sigma_i$, and apply the wedge product to the $\omega_i$ part.)