Is there such a thing as a weak coalgebra homomorphism?
The `weak morphisms' you describe do have a name in a slightly restricted setting. They can also be defined quite generally, as I'll explain.

In the case where $T : \mathsf{Set} \to \mathsf{Set}$ preserves weak pullbacks (many natural functors on $\mathsf{Set}$ do) it is known that behavioural equivalence coincides with coalgebraic bisimilarity. Then your morphisms are known as functional $\alpha$-step bisimulations where  $\alpha$ is an ordinal. Admittedly I've only ever seen them defined for ordinals $\alpha \leq \omega$. Prior to coalgebra, modal logicians had studied n-step bisimulations for Kripke frames, which amount to n-step bisimulations for coalgebras for the powerset functor. Your requirement that they be functions as opposed to relations makes them functional n-step bisimulations.

On the other hand, you can define the concept you are talking about in a far more general setting without referring to coalgebraic bisimulations. For any category $\mathbb{C}$ which has limits of ordinal-indexed cochains and any functor $T : \mathbb{C} \to \mathbb{C}$ one can define $T$ 's terminal sequence. The condition regarding limits is actually rather weak e.g. many categories (including $\mathsf{Set}$) are actually complete i.e. have all small limits. The terminal sequence is a diagram in $\mathbb{C}$ and looks like:

$1 \quad \xleftarrow{!_{T1}} \quad T1 \quad \xleftarrow{T !_{T1}} \quad T^2 1 \quad \xleftarrow{T^2 !_{T1}} \quad \dots \quad T^\omega 1  \quad \xleftarrow{f_\omega^{\omega + 1}} \quad T(T^\omega 1) \quad \xleftarrow{Tf_\omega^{\omega + 1}} \quad \dots$

Here $1$ is the terminal object in $\mathbb{C}$ (limit of empty cochain). For example in $\mathsf{Set}$ this can be taken to be the one element set $1 = \{*\}$. The map $!_{T1} : T1 \to 1$ is the unique morphism into the terminal object e.g. in $\mathsf{Set}$ is simply maps every element of $T1$ to $*$. Each $T^n 1$ is computed by iterating $T$ and $T^\omega 1$ is the limit of the cochain preceding it. One can then continue beyond $\omega$, if necessary. Intuitively $T^\alpha 1$ is the collection of  $\alpha$-step behaviours.

Now any $T$-coalgebra $(Z,\gamma)$ induces a cone over this sequence i.e.  a collection of $\mathbb{C}$-morphisms $\mathsf{beh}_\gamma^\alpha : Z \to T^\alpha 1$ for each ordinal $\alpha$. I'll just define them for $\alpha &lt; \omega$:

$\mathsf{beh}_\gamma^0 : Z \to 1$ is the unique map into the terminal object.

$\mathsf{beh}_\gamma^{n+1} = T\mathsf{beh}_\gamma^n \circ \gamma : Z \to T^{n+1} 1$

Intuitively these maps send a state in $Z$ to its $\alpha$-step behaviour. Now we can describe what you are talking about. Suppose we have two $T$-coalgebras $(A,\gamma)$ and $(B,\delta)$. Then a $\mathbb{C}$-morphism $f : A \to B$ preserves $\alpha$-step behaviour iff:

$\mathsf{beh}_\delta^\alpha \circ f = \mathsf{beh}_\gamma^\alpha$

That is, the $\alpha$-step behaviour of $f(z)$ in $\delta$ is precisely the $\alpha$-step behaviour of $z$ in $\gamma$.

Anyway, I hope this is helpful. You can find various references by googling 'terminal sequence coalgebra' or 'final sequence coalgebra'.