Raised index of partial derivative I am having a really hard time wrapping my head around component notation for tensor fields. For example, I do not know exactly what the following expression means
$$\partial_\mu\partial^\nu \phi, \tag{$\#$}$$ where $\phi$ is a scalar field. On the one hand $\partial^\nu=g^{\lambda\nu}\partial_\lambda$ where $g_{\mu\nu}$ is the Minkowski metric, and hence we could write explicitly $$\partial_\mu\partial^\nu \phi=\sum_{\mu,\nu,\lambda}g^{\lambda\nu}\partial_\mu \partial_{\lambda}\phi=\sum_{\mu,\nu}\partial_{\mu}\partial_{\nu}\phi=\partial_\mu\partial_\nu\phi. \tag{$*$} $$
On the other hand, we may think of $\partial_\mu\partial^\nu=g(\partial_\mu,\partial^\nu)=\delta_\mu^\nu;$ so that $\partial_\mu\partial^\nu\phi=\phi?$ Maybe? I am actually not sure of what this would mean. I am really confused. Any help is appreciated.
Edit:
To give context of where this expression comes from: I was computing the Lagrangian $$\mathcal L=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu \phi) $$ considering an infinitesimal spacetime translation $x^ \mu\to x^\mu-\alpha a^\mu$. The scalar field thus transforms like $\phi(x)\to \phi(x)+\alpha(\partial_\mu\phi(x))a^\mu.$
Plugging thins into the Lagrangian yields the term I am referring to.
Edit 2: The change in placement of indices are actually my doubts. I try to elaborate.
I do not have any background in using indices to talk about tensors. I am used to interpret the expressions $\partial_\mu$ as the local vector field defined in some chart (local coordinates). I think about vector fields $X$ as abstract section of the Tangent bundle, which restricted to local coordinates can be expressed as $X=X^\mu\partial_\mu$. In the context of QFT, as far as I understand, the symbol $\partial_\mu$ denotes  $(\partial_t,\nabla)$ in the local coordinates $(t,x,y,z)$. So that $\partial_\mu\phi=(\partial_t \phi,\partial_x \phi,\partial_y\phi,\partial_z\phi)$. This was supposed to be my justification on why I wrote the summation on $\mu$ and $\nu$ in $(*)$, but now I note that this only applies when $\mu$ or $\nu$ appear twice, indicating the scalar product; which leads me to the last remark. I think of $g_{\mu \nu}$ as the component of the matrix $$g=\begin{pmatrix}
1&0&0&0\\
0&-1&0&0\\
0&0&-1&0\\
0&0&0&-1\\
\end{pmatrix}$$
which represents the pseudo-Riemmanian metric, which by definition acts on tangent vectors, i.e. linear combinations of the $\partial_\mu$ applied to a point. This is where my doubt comes, in which was the right way to interpret the notation; in particular what is the expresion $(\#)$ in explicit coordinates?
 A: $\renewcommand{\lag}{\mathcal{L}}\renewcommand{\pd}{\partial}\renewcommand{\d}{\mathrm{d}}$$\pd^\mu$ is defined as $\pd^\mu := g^{\mu\nu}\pd_{\nu}$, where I use the convention that all repeated indices are summed and $g^{\mu\nu}$ are the components of the inverse metric tensor. Thus your Lagrangian can be rewritten as
$$\lag=\tfrac12g^{\mu\nu}(\pd_\mu\phi)(\pd_\nu\phi)\tag{1}$$
and also your expression $(\#)$ is equal to $g^{\mu\sigma}\pd_\nu\pd_\sigma\phi$.
To see where all this comes from from a differential geometry point of view, this Lagrangian can be written in a coordinate free form as the top-form
$$\lag = \tfrac12 \d\phi\wedge\star\d\phi,\tag{2}$$
where $\d$ is the exterior derivative and $\star$ is the Hodge-star. It is an easy exercise to restrict to a local coordinate system, $\d x^\mu$, in which case $\d\phi$ becomes $\frac{\pd\phi}{\pd x^\mu}\d x^\mu\equiv\pd_\mu\phi\,\d x^\mu$. The Hodge star will contribute a factor of $g^{\mu\nu}$ and so (2) will fall back to (1).
Moreover, you can think of $a^\mu\pd_\mu\phi(x)$ in a more formal setting as $\iota_a \d\phi$, where $\iota_a$ is the interior product along the vector field $a$ with components $a^\mu$. So the transformation $\phi(x)\mapsto\phi(x)+\alpha a^\mu \pd_\mu\phi(x)$ is written as
$$\phi(x)\mapsto \phi(x) + \alpha\,(\iota_a\d\phi)(x).$$
The relevant term in your expression ($\#$) comes from a term $\alpha \d\phi\wedge\star\d\iota_a \d\phi$ in the Lagrangian, basically it is just the $\alpha \star\d\iota_a \d\phi$ part. If we expand this in local coordinates $\{\d x^\sigma\}$, we get:
$$ \alpha \star\d\iota_a \d\phi = \alpha a^\mu \pd_\sigma\pd_\mu \phi\;\star\d x^\sigma  = \alpha a^\mu \pd_\sigma\pd_\mu \phi\ g^{\nu\sigma} \varepsilon_{\nu\lambda\kappa\rho}\d x^\lambda\wedge\d x^\kappa\wedge\d x^\rho,$$
where in the second equality I used the definition of the Hodge star acting on the basis differentials. Stripping off numbers, $\varepsilon$-symbols and the differentials, all we're left with is $$g^{\nu\sigma}\pd_\sigma\pd_\mu\phi,\tag{$\#'$}$$
which is exactly what you would have found (with your much shorter route) as
$$\pd^\nu\pd_\mu\phi \tag{#}.$$
Thus, $(\#')=(\#)$.
Of course the typical way to arrive there is to simply use the fact that for any object $\bullet_\mu$ with a downstairs leg we can lift it using the inverse metric, i.e. $\bullet^\mu := g^{\mu\nu}\bullet_\nu$. But since you had trouble understanding where does this stem from from a differential geometry perspective, I wanted to stick with the differential geometry picture all the way through, from the Lagrangian to the final result. Hope this helped and didn't confuse you more.
