Definition of the twisted spin covariant derivative for twisted spinor bundles Let $S\to M$ be the spinor bundle and
\begin{equation}
\nabla^S\colon\Omega^0(M,S)=\Gamma(M,S)\to\Omega^1(M,S)=\Gamma(M,T^*M\otimes S)
\end{equation}
the spin covariant derivative. In addition, let $E\to M$ be a vector bundle with a covariant derivative $\nabla^E\colon\Omega^0(M,E)\to\Omega^1(M,E).$ In some notes I'm reading the twisted spin covariant derivative (used to define the twisted Dirac operator) is defined as
\begin{equation}\tag{1}
\nabla:=\nabla^S\otimes 1+1\otimes\nabla^E.
\end{equation}
However, I don't see how the operators $\nabla^S\otimes 1$ and $1\otimes\nabla^E$ are defined, since we can't write
\begin{equation}
\Omega^0(M,S\otimes E)\cong\Omega^0(M,S)\otimes E\cong S\otimes\Omega^0(M,E)
\end{equation}
(these expressions are undefined), unlike for vector valued functions, where
\begin{equation}
C^\infty(M,V\otimes W)\cong C^\infty(M,V)\otimes W\cong V\otimes C^\infty(M,W).
\end{equation}
Edit: What about
\begin{equation}
\Omega^0(M,S\otimes E)\cong\Omega^0(M,S)\otimes\Omega^0(M,E)
\end{equation}
and
\begin{equation}
\Omega^1(M,S\otimes E)\cong\Omega^1(M,S)\otimes\Omega^0(M,E)\cong\Omega^0(M,S)\otimes\Omega^1(M,E)
\end{equation}
If that's correct, I guess I see how $(1)$ would make sense.
 A: Not every section of $S\otimes E$ is of the form $s\otimes e$ with $s\in \Gamma(S)$ and $e\in \Gamma(E)$. But it locally is a finite sum of such tensor product of sections. That is, if $\gamma \in \Gamma(S\otimes E)$, then, locally:
$$
\gamma = \sum_{i=1}^N s_i\otimes e_i
$$
for $s_i$ local sections of $S$ and $e_i$ local sections of $E$. Hence, we define $\nabla$ as:
$$
\nabla \gamma = \sum_{i=1}^N \left(\nabla^Ss_i\right)\otimes e_i + s_i\otimes \left(\nabla^E e_i\right).
$$
One can show this does not depend on the representatives chosen in a local representation.
In short : $\Gamma(S\otimes E)$ is locally generated by simple tensors $s\otimes e$.
Comment In the comments section, there is a "problem": let $f$ be a smooth function and $s \in \Gamma(S)$, $e\in \Gamma(E)$. Write $\tilde{s} = fs$ and $\tilde{e} = fe$. Then $\tilde{s}\otimes e = s \otimes \tilde{e}$, and:
\begin{align}
\nabla\left(\tilde{s}\otimes e\right) &= \left(\nabla^S\tilde{s} \right)\otimes e + \tilde{s}\otimes \nabla^E e \\
&= \left(\mathrm{d}f \otimes s + f\nabla^Ss \right)\otimes e + f s\otimes \nabla^E e \\
&= \mathrm{d}f\otimes s \otimes e + f \left(\nabla^Ss\otimes e + s\otimes\nabla^Ee \right)\\
&= \mathrm{d}f \otimes \left( s\otimes e\right) + f \nabla \left(s\otimes e\right).
\end{align}
The same computations with $\tilde{e}$ show that $\nabla \left(f s\otimes e \right)$ does not depend on where we put $f$ as a coefficient.
