**Update: A counterexample.** I've realized that the claim you are trying to generalize from sets to classes is not actually true for sets. Specifically, there are directed well-quasi-ordered sets that do not admit any linearly ordered cofinal subset. And a similar counterexample exists for classes.

**Theorem.** There is a directed well-quasi-order set having no
cofinal well-ordered subset.

Proof. Consider the order on $\omega_1\times\omega$, where
$(\alpha,n)\leq(\alpha',n')$ just in case $\alpha\leq\alpha'$ and
$n\leq n'$. It is easy to see that this is well-founded and
directed. Furthermore, it is well-quasi-ordered, because suppose
that $A$ is an infinite antichain in $\Gamma$. Let $(\alpha,n)$
have the smallest value of $n$ arising from a pair in $A$. Thus,
all other pairs $(\beta,m)$ in $A$ have $\beta\lt\alpha$ and $m\gt
n$. And as $m$ goes up, the $\beta$'s must go down, which is
impossible.

But I claim that there is no well-ordered cofinal suborder of
$\Gamma$. The reason is the mis-match of the cofinality on the two
coordinates. If $\langle \alpha_\xi,n_\xi\rangle$ for
$\xi\lt\delta$ is cofinal and increasing in the order, then it
must be that $\{n_\xi\mid\xi\lt\delta\}$ is cofinal in $\omega$
and also $\{\alpha_\xi\mid\xi\lt\delta\}$ is cofinal in
$\omega_1$. So $\delta$ would have to have cofinality $\omega$ and
also $\omega_1$, a contradiction. QED

The conclusion is that it is not generally possible to find cofinal
linearly ordered suborders of well quasi orders, even when they
are sets.

A slightly easier example shows the flaw in your construction. Consider the order
on $\mathbb{N}\times\mathbb{N}$ for which $(n,m)\leq (n',m')$ when
both $n\leq n'$ and $m\leq m'$. This is a well quasi-order, since
it is well-founded and there is no infinite antichain, since for a
given point $(n,m)$, any point incomparable must either have
smaller $n$ or smaller $m$, but any two points with one coordinate
the same are comparable; so any antichain containing $(n,m)$ has
at most $n+m+1$ many elements. Also, the order is clearly upward
directed. But observe that your procedure, depending on $\phi$, could easily
have led you to choose $(0,0), (1,0), (2,0), (3,0),\ldots$ as the
initial segment of your sequence. In this case, you would be
stuck, since this is an increasing sequence in the order, but it
is not cofinal---since nothing in it is above $(3,3)$---but
meanwhile, there is no way to continue the sequence, since the
sequence also is not dominated in the order; there is nothing to
put at stage $\omega$. (Meanwhile, the order does have a cofinal $\omega$-sequence,
namely, $(0,0),(1,1),(2,2),\ldots$. The point is that the
procedure you describe in your proof doesn't necessarily find it.)

One may modify the counterexample to proper classes, to give a counterexample to your question about proper classes, as follows. Consider the similar class order on $\text{Ord}\times\omega$, where $(\alpha,n)\leq(\alpha',n')$ if and only if $\alpha\leq\alpha'$ and $n\leq n'$. This is directed and well-quasi-ordered, just as the order above, but there is no cofinal order-preserving map $\xi\mapsto (\alpha_\xi,n_\xi)$ from $\text{Ord}$ into it, since the values of $n_\xi$ must eventually increase, and so there will be a stage $\beta$ for which $\sup_{\xi\lt\beta}n_\xi=\omega$, after which time the sequence cannot continue. This is essentially the same kind of counterexample as above, resulting from the mismatched cofinalities of $\text{Ord}$ and $\omega$. Meanwhile, there is a surjective map $\text{Ord}\to\text{Ord}\times\omega$, whose image is therefore cofinal, by using pairing functions on the ordinals, and therefore your extra hypothesis is also fulfilled for this counterexample.

**Original answer.** In general, you cannot get an order-preserving cofinal map from Ord to a class order $\Gamma$, even when $\Gamma$ is fully well-ordered. The reason is that the cofinality of $\Gamma$ may be strictly less than Ord.

For example, consider the order $\Gamma=\text{Ord}\cdot\omega$, meaning $\omega$ many copies of Ord, one on top of another. You can build this on $\omega\times\text{Ord}$ with the lexical order. This is a well-order (and hence a quasi well order), in the sense that every subclass has a least member, but since it has a cofinal $\omega$ sequence, there can be no order-preserving cofinal map from Ord to $\Gamma$, since this would give a cofinal $\omega$ sequence in Ord, contrary to the replacement axiom. Indeed, there is also no order-preserving cofinal map from $\omega_1$ or indeed, from any uncountable regular cardinal, into $\text{Ord}\cdot\omega$.

The essential issue here is that this $\Gamma$ is not *set-like*, where a relation is *set-like* when the sets of predecessors of any point form a set. For a set-like proper class well-ordering $\Lambda$, your construction from the set case adapts to give you an order-preserving cofinal map from Ord to $\Lambda$, for in this case, as you construct the cofinal Ord sequence, at each ordinal stage the sequence you have produced so far will not yet be cofinal, and so you can continue.

Let me now give an example showing that even in the case the quasi-well-order is set-like, one nevertheless needs global choice to find the embedding, and so it not possible to do it merely in GB+AC. To see this, consider one of the standard models of GB+AC + $\neg$ global choice. For example, such a model is constructed in my answer to Asaf Karagila's question, Does ZFC prove the universe is linearly orderable?. The corollary in that argument can be strengthed to the following claim, proved by the same argument.

**Theorem.** Every model of ZFC has a class forcing extension to a model of GB+AC, in which there is a class of pairs, such that no class makes proper-class-many choices from those pairs.

Thus, in this model we have a class $\{\{A_\alpha,B_\alpha\}\mid \alpha\in\text{Ord}\}$, such that no class makes a choice for unboundedly many $\alpha$. We may now form the quasi-well-order $\Gamma$ that puts the pairs as antichains in a tower of height $\text{Ord}$, but there can be no order-preserving cofinal embedding of $\text{Ord}$ into $\Gamma$ in this model, since such an embedding would make unboundedly many choices from the family.

Conclusion: one really needs global choice, and not just ZFC or even GB+AC, even when the relation is set-like.

2more comments