1. No, one of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is not correct if it is supposed to mean partial derivatives. The correct Euler-Lagrange (EL) equations read $$ \tag{2'} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} - \sum_{\mu} \color{Red}{\frac{ d}{dx^{\mu}}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi^{\alpha})} + \ldots,$$ where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher derivative terms. Here $$ \color{Red}{\frac{ d}{dx^{\mu}}}~=~ \frac{\partial }{\partial x^{\mu}} +\sum_{\alpha}(\partial_{\mu}\phi^{\alpha})\frac{\partial }{\partial \phi^{\alpha}} + \sum_{\alpha, \nu} (\partial_{\mu}\partial_{\nu}\phi^{\alpha})\frac{\partial }{\partial (\partial_{\nu}\phi^{\alpha})} + \ldots $$ is the $\color{Red}{\text{total spacetime derivative}}$ rather than a partial spacetime derivative. See also this and this related Phys.SE posts.

2. Let us mention for completeness that the other appearance of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is correct. It may be replaced with a total spacetime derivative $\color{Red}{d_{\mu}}$, since $\partial_{\mu}\phi\equiv\color{Red}{d_{\mu}}\phi$ by definition, cf. OP's eq. (3).

1. No, one of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is not correct if it is supposed to mean partial derivatives. The correct Euler-Lagrange (EL) equations read $$ \tag{2'} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} - \sum_{\mu} \color{Red}{\frac{ d}{dx^{\mu}}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi^{\alpha})} + \ldots,$$ where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher derivative terms. Here $$ \color{Red}{\frac{ d}{dx^{\mu}}}~=~ \frac{\partial }{\partial x^{\mu}} +\sum_{\alpha}(\partial_{\mu}\phi^{\alpha})\frac{\partial }{\partial \phi^{\alpha}} + \sum_{\alpha, \nu} (\partial_{\mu}\partial_{\nu}\phi^{\alpha})\frac{\partial }{\partial (\partial_{\nu}\phi^{\alpha})} + \ldots $$ is the $\color{Red}{\text{total spacetime derivative}}$ rather than a partial spacetime derivative. See also this and this related Phys.SE posts.

2. Let us mention for completeness that the other appearance of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is correct. It may be replaced with a total spacetime derivative $\color{Red}{d_{\mu}}$, since $\partial_{\mu}\phi\equiv\color{Red}{d_{\mu}}\phi$ by definition, cf. OP's eq. (3).

No, one of the partial derivative symbol in OP's equation (2) is not correct if it is supposed to mean partial derivatives. The correct Euler-Lagrange (EL) equations read $$ \tag{2'} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} - \sum_{\mu} \color{Red}{\frac{ d}{dx^{\mu}}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi^{\alpha})} + \ldots,$$ where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher derivative terms. Here

$$ \color{Red}{\frac{ d}{dx^{\mu}}}~=~ \frac{\partial }{\partial x^{\mu}} +\sum_{\alpha}(\partial_{\mu}\phi^{\alpha})\frac{\partial }{\partial \phi^{\alpha}} + \sum_{\alpha, \nu} (\partial_{\mu}\partial_{\nu}\phi^{\alpha})\frac{\partial }{\partial (\partial_{\nu}\phi^{\alpha})} + \ldots $$

is the $\color{Red}{\text{total spacetime derivative}}$ rather than a partial spacetime derivative. See also this and this related Phys.SE posts.

1. No, one of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is not correct if it is supposed to mean partial derivatives. The correct Euler-Lagrange (EL) equations read $$ \tag{2'} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} - \sum_{\mu} \color{Red}{\frac{ d}{dx^{\mu}}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi^{\alpha})} + \ldots,$$ where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher derivative terms. Here $$ \color{Red}{\frac{ d}{dx^{\mu}}}~=~ \frac{\partial }{\partial x^{\mu}} +\sum_{\alpha}(\partial_{\mu}\phi^{\alpha})\frac{\partial }{\partial \phi^{\alpha}} + \sum_{\alpha, \nu} (\partial_{\mu}\partial_{\nu}\phi^{\alpha})\frac{\partial }{\partial (\partial_{\nu}\phi^{\alpha})} + \ldots $$ is the $\color{Red}{\text{total spacetime derivative}}$ rather than a partial spacetime derivative. See also this and this related Phys.SE posts.

2. Let us mention for completeness that the other appearance of the partial derivative symbol $\partial_{\mu}$ in OP's equation (2) is correct. It may be replaced with a total spacetime derivative $\color{Red}{d_{\mu}}$, since $\partial_{\mu}\phi\equiv\color{Red}{d_{\mu}}\phi$ by definition, cf. OP's eq. (3).