Let $\alpha$ and $\beta^p$ be respectively $0$ and $p$ vector valued forms. Then $\alpha\cdot \beta^p$ is a scalar valued $p$ form. How is
$$ \begin{equation} \mathrm{d}\left(\alpha\cdot \beta^p \right) \end{equation} $$
defined in terms of $\mathrm{d}\alpha$ and $\mathrm{d}\beta^p$ ? Here $\cdot$ is the standard scalar product (assuming that these are vectors in $\mathbb{R}^3$ but maybe it's possible to generalize). Moreover, how do we calculate
$$ \begin{equation} \left(\alpha\cdot \beta^p \right)\wedge\left(\gamma\cdot \delta^q \right) \end{equation} $$
where $\gamma$ and $\delta^q$ are respectively $0$ and $q$ vector valued forms ?
I've tried simply expanding in vector components in a general basis but it's no so clean notationally and in the end it cannot be written a single dot product.
As far as I understand, $\boldsymbol{\alpha}$ is a vector valued $0$-form, in other words, a vector valued function, or, a vector field: $$ \boldsymbol{\alpha}=\alpha^\rho\boldsymbol{e}_\rho $$ where the summation convention is used. Further, $\boldsymbol{\beta}$ is a vector valued $p$-form: $$ \boldsymbol{\beta}={\beta^\rho}_{|\mu_1...\mu_p|}\boldsymbol{d}x^{\mu_1}\,\wedge...\wedge \,\boldsymbol{d}x^{\mu_p} $$ where the summation convention is used and $|\mu_1...\mu_p|$ stands for index combinations with $\mu_1<...<\mu_p\,.$
When you say that $\boldsymbol{\alpha}\cdot\boldsymbol{\beta}$ is a scalar valued $p$-form it seems you mean
$$ \boldsymbol{\alpha}\cdot\boldsymbol{\beta}=\underbrace{\alpha_\rho\,{\beta^\rho}_{|\mu_1...\mu_p|}}_{\text{scalar}}\,\boldsymbol{d}x^{\mu_1}\wedge...\wedge \boldsymbol{d}x^{\mu_p}\,. $$ The exterior derivative of this should then be (as always) $$ \boldsymbol{d}(\boldsymbol{\alpha}\cdot\boldsymbol{\beta})=\partial_\sigma(\alpha_\rho\,{\beta^\rho}_{|\mu_1...\mu_p|})\,\boldsymbol{d}x^\sigma\wedge \boldsymbol{d}x^{\mu_1}\wedge...\wedge \boldsymbol{d}x^{\mu_p}\,. $$ You are asking how this can be expressed in terms of $\boldsymbol{d}\boldsymbol{\alpha}$ and $\boldsymbol{d}\boldsymbol{\beta}\,.$ Clearly, $\boldsymbol{d}\boldsymbol{\alpha}$ is a vector valued $1$-form. In chapter 14 of [1] we find the expressions \begin{align} \boldsymbol{d}\boldsymbol{\alpha}&=\boldsymbol{e}_\rho\boldsymbol{d}\alpha^\rho +\alpha^\rho \boldsymbol{de}_\rho\,,\\ \boldsymbol{de}_\rho&=\boldsymbol{e}_\nu{\boldsymbol{\omega}^\nu}_\rho \end{align} where the coefficients ${\boldsymbol{\omega}^\nu}_\rho$ by which the vector valued $1$-form $\boldsymbol{de}_\rho$ is expanded are closely related to curvature, i.e., to the difference of the partial derivative from the covariant derivative. As far as I can tell we will get a nice result when we assume that all $\boldsymbol{\omega}$ vanish. Then $$ \boldsymbol{d}\boldsymbol{\alpha}=\boldsymbol{e}_\rho\boldsymbol{d}\alpha^\rho=\boldsymbol{e}_\rho(\partial_\sigma\alpha^\rho)\boldsymbol{d}x^\sigma\,. $$ since each $\alpha^\rho$ is a scalar.
Handling the vector valued $(p+1)$-form $\boldsymbol{d\beta}$ is simpler: $$ \boldsymbol{d}\boldsymbol{\beta}=\partial_\sigma({\beta^\rho}_{|\mu_1...\mu_p|})\,\boldsymbol{d}x^\sigma\wedge \boldsymbol{d}x^{\mu_1}\wedge...\wedge \boldsymbol{d}x^{\mu_p}\,. $$ It is also clear that $$ \partial_\sigma(\alpha_\rho\,{\beta^\rho}_{|\mu_1...\mu_p|})=(\partial_\sigma\alpha_\rho){\beta^\rho}_{|\mu_1...\mu_p|}+\alpha_\rho(\partial_\sigma {\beta^\rho}_{|\mu_1...\mu_p|})\,. $$ Thus, if there is no curvature we should get the not so surprising result $$ \boldsymbol{d}(\boldsymbol{\alpha}\cdot\boldsymbol{\beta})=\boldsymbol{\alpha}\cdot\boldsymbol{d\beta}+\boldsymbol{d\alpha}\cdot\boldsymbol{\beta}\,. $$
[1] Misner, Thorne & Wheeler, Gravitation.