If you are confused by the Einstein summation convention as well as the use of the metric $g_{\mu\nu}$ and its inverse $g^{\mu\nu}$ to move up and down indices, you can write everything without using those. When you do this, each greek letter $\mu,\nu,\ldots$ which appear is an integer labeling the $4$ coordinates on your $4$-dimensional space time.
So, if you call your coordinates $x^0 = t$, $x^1 = x$, $x^2 = y$ and $x^3 = z$, then $\partial_0 \varphi = \frac{\partial \phi}{\partial x^0} = \frac{\partial \phi}{\partial t}$ and likewise for $\mu =1,2,3$. Therefore, we have, for any values of $\mu$ and $\nu$ :
$$\frac{\partial}{\partial (\partial_\mu \phi)} \big[\partial_\nu \phi\big] = \delta^\mu_\nu= \left\{\begin{array}{cc} 1 &\text{if } \mu = \nu \\ 0 & \text{if } \mu\neq \nu \end{array}\right.$$
The Klein-Gordon Lagrangian (with $m=0$ for simplicity) is :
$$L = \sum_{\mu = 0}^4\sum_{\nu = 0}^4 \frac 12 \frac{\partial \phi}{\partial x^\mu} \frac{\partial \phi}{\partial x^\nu} g^{\mu\nu} $$
If you want to compute $\frac{\partial L}{\partial (\partial_\alpha\varphi)}$, for any given value of $\alpha$, you have :
\begin{align}
\frac{\partial L}{\partial (\partial_\alpha\varphi)} &= \frac{\partial }{\partial (\partial_\alpha\varphi)}\left[\sum_{\mu = 0}^4\sum_{\nu = 0}^4 \frac 12 \frac{\partial \phi}{\partial x^\mu} \frac{\partial \phi}{\partial x^\nu} g^{\mu\nu}\right] \\
&=\frac 12\sum_{\mu = 0}^4\sum_{\nu = 0}^4 \left( \left[\frac{\partial }{\partial (\partial_\alpha\varphi)}\frac{\partial \phi}{\partial x^\mu}\right] \frac{\partial \phi}{\partial x^\nu} g^{\mu\nu} + \frac{\partial \phi}{\partial x^\mu} \left[\frac{\partial }{\partial (\partial_\alpha\varphi)}\frac{\partial \phi}{\partial x^\nu}\right] g^{\mu\nu}\right)\\
&= \frac 12 \left(\sum_{\nu = 0}^4 \frac{\partial \phi}{\partial x^\nu} g^{\alpha\nu} + \sum_{\mu = 0}^4\frac{\partial \phi}{\partial x^\mu} g^{\mu \alpha}\right) \\
&= \sum_{\mu = 0}^4g^{\alpha \mu}\frac{\partial \phi}{\partial x^\mu}
\end{align}
where on the last line we used the fact that the metric is symmetric and that we can rename dummy indices.
Rewriting this with the Einstein summation convention we get :
$$\frac{\partial L}{\partial (\partial_\alpha \phi)} = g^{\alpha\mu}\partial_\mu\phi$$
and if we raise the indices using the (inverse) metric, we get :
$$\frac{\partial L}{\partial (\partial_\alpha \phi)} = \partial^\alpha \phi$$
