It is somewhat counterintuitive to want tetration to be right-associative over ordinals when ordinal arithmetic is left-associative. To merge these two, I simply took the supremum of the left- and right-associative approaches. Define the following operations on ordinals:
$$\alpha\uparrow^\mu\beta:=\begin{cases}\alpha\cdot\beta,&\mu=0\\1,&\mu\ne0\land\beta=0\\\sup\{\alpha\uparrow^{\mu'}(\alpha\uparrow^\mu\beta'),(\alpha\uparrow^\mu\beta')\uparrow^{\mu'}\alpha,~|~\beta'\in\beta\land\mu'\in\mu\},&\mu\ne0\land\beta\ne0\end{cases}$$
which gives the standard $\alpha^\beta$ by $\alpha\uparrow^1\beta$ and generalizes this neatly, at least for $\alpha\ge\omega$. It is clear that this doesn't stop at fixed-points such as $\omega^{\varepsilon_0}=\varepsilon_0$, and it gives nice identities such as $\omega\uparrow^2\omega=\varepsilon_0$, which intuitively seems to be good.
In general, what is $\alpha\uparrow^\mu\beta$ in Veblen normal form with all arguments written in this normal form inductively, for $\mu,\beta<\Gamma_0$, the Feferman–Schütte ordinal?
To start off with the basics, note that
$$\omega\uparrow^21=\omega\\\omega\uparrow^22=\omega^\omega\\\omega\uparrow^23=\omega^{\omega^\omega}\\\vdots\\\omega\uparrow^2\omega=\varepsilon_0=\varphi_1(0)\\\omega\uparrow^2(\omega+1)=\varepsilon_0^\omega=\omega^{\omega^{\varepsilon_0+1}}$$
In general, $\omega\uparrow^2(\omega\cdot(1+\beta))=\varepsilon_\beta$, and more generally, $\varepsilon_\alpha\uparrow^2(1+\omega\cdot\beta)=\varepsilon_{\alpha+\beta}=\varphi_1(\alpha+\beta)$, for $\beta>0$. We also have $\varepsilon_\alpha\uparrow^2(1+\omega\cdot\beta+1)=\omega^{\omega^{\varepsilon_{\alpha+\beta}+1}}$, and from there $\varepsilon_\alpha\uparrow^2(1+\omega\cdot\beta+k)$ is just $(k-1)$ more $(\omega\widehat~)$'s.
In the same way,
$$\varphi_\mu(\alpha)\uparrow^{1+\mu}(\omega\cdot\beta)=\varphi_\mu(\alpha+\beta)\tag0$$
and
$$\varphi_\mu(\alpha)\uparrow^{1+\mu}(\omega\cdot\beta+k)=\sup_{\mu'\in\mu}\varphi_{\mu'}^k(\varphi_\mu(\alpha+\beta)+1)\tag1$$
for all $\beta>0$ and $\mu>2$ and $0<k<\omega$.
Proof:
It is clear these hold for the base cases. We show the inductive step:
It is somewhat straightforward to show from the definition that $\alpha\uparrow^\mu\beta$ is continuous in $\beta$, which allows us to pass limits through the last argument.
It is also somewhat clear that this is increasing in all arguments, which gives us flexibility when taking limits.
Suppose that $(0)$ and $(1)$ holds for all $\alpha$ and $2<\mu<\mu^\star$ and $\beta>0$. Suppose that $(0)$ and $(1)$ holds for all $\alpha$ and $\mu=\mu^\star$ and $0<\beta<\beta^\star$. We show this holds for $\mu=\mu^\star$ and $\beta=\beta^\star$:
Suppose that $\beta^\star=\omega\cdot\beta'$ for successor $\beta'=\gamma+1$ is a non-zero limit ordinal. Then we have
\begin{align}\varphi_\mu(\alpha)\uparrow^{1+\mu}\beta^\star&=\sup_{n\in\mathbb N}\varphi_\mu(\alpha)\uparrow^{1+\mu}(\omega\cdot\gamma+n)\\&=\sup_{n\in\mathbb N,\mu'\in\mu}\varphi_{\mu'}^n(\varphi_\mu(\alpha+\gamma)+1)\\&=\varphi_\mu(\alpha+\gamma+1)\\&=\varphi_\mu(\alpha+\beta')\end{align}
Suppose that $\beta^\star=\omega\cdot\beta'$ for non-zero limit $\beta'$ is a non-zero limit ordinal. Then we have
\begin{align}\varphi_\mu(\alpha)\uparrow^{1+\mu}\beta^\star&=\sup_{\gamma\in\beta'}\varphi_\mu(\alpha)\uparrow^{1+\mu}(\omega\cdot\gamma)\\&=\sup_{\gamma\in\beta'}\varphi_\mu(\alpha+\gamma)\\&=\varphi_\mu(\alpha+\beta')\end{align}
Suppose that $\beta^\star=\beta'+k$ for non-zero limit $\beta'=\omega\cdot\gamma$ and $1<k<\omega$ is the $k$th successor of a limit ordinal. Then we have
\begin{align}\varphi_\mu(\alpha)\uparrow^{1+\mu}\beta^\star&=\sup_{\mu'\in\mu}(\varphi_\mu(\alpha)\uparrow^{1+\mu}(\omega\cdot\gamma+k-1))\uparrow^{1+\mu'}\varphi_\mu(\alpha)\\&=\sup_{\mu'\in\mu}\varphi_{\mu'}^{k-1}(\varphi_\mu(\alpha+\gamma)+1)\uparrow^{1+\mu'}\varphi_\mu(\alpha)\\&=\sup_{\mu'\in\mu}\varphi_{\mu'}^k(\varphi_\mu(\alpha+\gamma)+1)\uparrow^{1+\mu'}(\omega\cdot\varphi_\mu(\alpha))\\&=\sup_{\mu'\in\mu}\varphi_{\mu'}^k(\varphi_\mu(\alpha+\gamma)+1+\varphi_\mu(\alpha))\\&=\sup_{\mu'\in\mu}\varphi_{\mu'}^k(\varphi_\mu(\alpha+\gamma)+1)\end{align}
Q.E.D.
From there we can see that to find $\alpha\uparrow^\mu\beta$ for limit $\beta$, it suffices to round $\alpha$ down to the nearest Veblen function and go from there.