We have that
\begin{equation}
\tag{$\ast$}
d(f \cap g)
= (-1)^n d(f) \cap g + (-1)^n f \cap d(g)
= (-1)^{|f|} \Bigl( d(f) \cap g + f \cap d(g) \Bigr).
\end{equation}
for all $f \in \operatorname{C}^n(A) = \operatorname{Hom}_k(A^{\otimes n}, A)$ and $g \in \operatorname{C}_p(A, M) = M \otimes A^{\otimes p}$ (the computation is at the end of this post).
From what I understand this may or may not be a problem, depending on what we are planning to do:
The equation $(\ast)$ sufficies to show that the cup product descends to a well-defined bilinear map $\operatorname{HH}^\bullet(A) \times \operatorname{HH}_\bullet(A,M) \to \operatorname{HH}_\bullet(A,M)$, (nearly) making $\operatorname{HH}_\bullet(A,M)$ into a graded $\operatorname{HH}^\bullet(A)$-module as claimed in Witherspoon’s text (see below for a possible problem).
From what I understand the Hochschild chain complex $\operatorname{C}_\bullet(A,M)$ does not become a dg $\operatorname{C}^\bullet(A)$-module as we would instead need the sign convention
\begin{equation*}
\tag{$\ast\ast$}
d(f \cap g) = d(f) \cap g + (-1)^n f \cap d(g).
\end{equation*}
But I need to point out that there are some things about the cap product which confuse me, and the correct understanding of which may show that there are actually no problems.
It seems that both Witherspoon and Belmans try to define a left module structure on $\operatorname{C}_\bullet(A,M)$ and $\operatorname{HH}_\bullet(A,M)$ via the cap product (where $\operatorname{C}^\bullet(A)$ and $\operatorname{HH}^\bullet(A)$ are equipped with the cup product).
But it seem to me that
$$
(f_1 \cup f_2) \cap g
= f_2 \cap (f_1 \cap g),
$$
so that the cap product actually defines a right module structure.
(Maybe this is no problem for the Hochschild (co)homology because the cup product is graded-commutative on $\operatorname{HH}^\bullet(A)$?)
Because $\operatorname{C}^\bullet(A)$ is a cochain complex but $\operatorname{C}_\bullet(A,M)$ is a chain complex there might be some more signs hidden in the definition of an dg module (as both differentials go in different directions).
I suspect that we want that the flipped complex $\widetilde{\operatorname{C}}^\bullet(A,M) = \operatorname{C}_{-\bullet}(A,M)$ to be dg $\operatorname{C}^\bullet(A)$-module with the sign convention as in $(\ast\ast)$, which would then mean that we always want the sign convention $(\ast\ast)$.
Now the calculations:
To show $(\ast)$ we start by computing the terms $d(f \cap g)$, $f \cap d(g)$ and $d(f) \cap g$.
For $n \leq p$ all three terms are zero, so let’s only look at $n > p$.
Let’s also assume that $g$ is a pure tensor $g = m \otimes a_1 \otimes \dotsb \otimes a_p$.
First term:
\begin{align*}
&\,
d( f \cap (m \otimes a_1 \otimes \dotsb \otimes a_p ) )
\\
=&\,
(-1)^n d( m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_p )
\\
=&\, (-1)^n \Big[
m f(a_1 \otimes \dotsb \otimes a_n) a_{n+1} \otimes a_{n+2} \otimes \dotsb \otimes a_p
\\
&\, \phantom{(-1)^n \Big[}
+ \sum_{i=1}^{p-n-1} (-1)^i
m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_{n+i} a_{n+i+1} \otimes \dotsb \otimes a_p \\
&\, \phantom{(-1)^n \Big[}
+ (-1)^{p-n} a_p m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_{p-1}
\Big]
\\
=&\, (-1)^n m f(a_1 \otimes \dotsb \otimes a_n) a_{n+1} \otimes a_{n+2} \otimes \dotsb \otimes a_p
\tag{1}
\\
&\, + (-1)^n \sum_{i=1}^{p-n-1} (-1)^i
m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_{n+i} a_{n+i+1} \otimes \dotsb \otimes a_p
\tag{2}
\\
&\, + (-1)^p a_p m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_{p-1}
\tag{3}
\end{align*}
Second Term:
\begin{align*}
&\,
d(f) \cap (m \otimes a_1 \otimes \dotsb \otimes a_p)
\\
=&\,
(-1)^{n+1} m \, d(f)(a_1 \otimes \dotsb \otimes a_{n+1}) \otimes a_{n+2} \otimes \dotsb \otimes a_p
\\
=&\,
(-1)^{n+1} m
\Big[
a_1 f (a_2 \otimes \dotsb \otimes a_{n+1})
\\
&\, \phantom{(-1)^{n+1} m \Big[}
+ \sum_{i=1}^n (-1)^i f(a_1 \otimes \dotsb \otimes a_i a_{i+1} \otimes \dotsb \otimes a_{n+1})
\\
&\, \phantom{(-1)^{n+1} m \Big[}
+ (-1)^{n+1} f(a_1 \otimes \dotsb \otimes a_n) a_{n+1}
\Big]
\otimes a_{n+2} \otimes \dotsb \otimes a_p
\\
=&\,
(-1)^{n+1} m a_1 f(a_2 \otimes \dotsb \otimes a_{n+1}) \otimes a_{n+2} \otimes \dotsb \otimes a_p
\tag{4}
\\
&\,
+ (-1)^{n+1} \sum_{i=1}^n (-1)^i
m f(a_1 \otimes \dotsb \otimes a_i a_{i+1} \otimes a_{n+1}) \otimes a_{n+2} \otimes \dotsb \otimes a_p
\tag{5}
\\
&\,
+ m f(a_1 \otimes \dotsb \otimes a_n) a_{n+1} \otimes a_{n+2} \otimes \dotsb \otimes a_p
\tag{1}
\end{align*}
Third term:
\begin{align*}
&\,
f \cap d(m \otimes a_1 \otimes \dotsb \otimes a_p)
\\
=&\, f \cap \Big[
m a_1 \otimes a_2 \otimes \dotsb \otimes a_p
\\
&\, \phantom{f \cap \Big[}
+ \sum_{i=1}^{p-1} (-1)^i m \otimes a_1 \otimes \dotsb \otimes a_i a_{i+1} \otimes \dotsb \otimes a_p
\\
&\, \phantom{f \cap \Big[}
+ (-1)^p a_p m \otimes a_1 \otimes \dotsb \otimes a_{p-1}
\Big]
\\
=&\,
(-1)^n m a_1 f(a_2 \otimes \dotsb \otimes a_{n+1}) \otimes a_{n+2} \otimes \dotsb \otimes a_p
\tag{4}
\\
&\, + (-1)^n \sum_{i=1}^n (-1)^i
m f(a_1 \otimes \dotsb \otimes a_i a_{i+1} \otimes \dotsb \otimes a_{n+1}) \otimes a_{n+2} \otimes \dotsb \otimes a_p
\tag{5}
\\
&\, + (-1)^n \sum_{i=1}^{p-n-1} (-1)^{n+i}
m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_{n+i} a_{n+i+1} \otimes \dotsb \otimes a_p
\tag{2}
\\
&\, + (-1)^{n+p} a_p m f(a_1 \otimes \dotsb \otimes a_n) \otimes a_{n+1} \otimes \dotsb \otimes a_{p-1}
\tag{3}
\end{align*}
We now want to combine these three terms in an equation of the form
$$
d(f \cap g) = \varepsilon_1 d(f) \cap g + \varepsilon_2 f \cap d(g)
$$
for suitable signs $\varepsilon_1, \varepsilon_2 = \pm 1$, which may depend on the degrees $n$ and $p$.
We get from $(1)$ that $\varepsilon_1 = (-1)^n$ and from $(2)$ that $\varepsilon_2 = (-1)^n$. This is also compatible with $(3)$, and the additional terms of the forms $(4)$ and $(5)$ cancel each other out because $\varepsilon_1 = \varepsilon_2$.
We thus find that
$$
d(f \cap g)
= (-1)^n d(f) \cap g + (-1)^n f \cap d(g)
= (-1)^{|f|} \Bigl( d(f) \cap g + f \cap d(g) \Bigr).
$$
