It is an exercise problem in Kunen (VII G5). I shall show that $\operatorname{Fn}(\lambda, 2, \lambda)$ adds a map from $\theta = \operatorname{cf}\lambda$ onto $\lambda^+$ for singular $\lambda$. The hint of the book says: At first show that the forcing poset adds a surjection from $\theta$ to $\lambda$. Now prove that $\bigcup G$ codes every function in $V$ from $\theta$ into $\lambda$. I proved the first one but could not prove the second one.
My attempt is: regard $\operatorname{Fn}(\lambda, 2,\lambda)$ as $\operatorname{Fn}(\lambda\times \theta\times\theta, 2,\lambda)$. In $V[G]$, we can find an one-to-one correspondence $h$ between $\theta$ and $\lambda$ and it has a name $\tau$. For $q\in G$ in the forcing poset satisfying $$q\Vdash \tau\text{ is an one-to-one correspondence between $\check{\theta}$ and $\check{\lambda}$}$$ and $x\in ({^\theta}\lambda)^V$, define $$ \begin{align} D_x &= \{p\mid \exists \xi<\lambda : \{\xi\}\times\theta\times\theta\subseteq \operatorname{dom}p \land \\ &\qquad\forall \mu,\nu<\theta: p(\xi,\mu,\nu)=1\iff q\Vdash \exists\sigma:\sigma=\check{x} \land \sigma(\check{\mu}) = \tau(\check{\nu})\} \end{align}$$
If we prove $D_x$ is dense (or dense below $q$), then $\bigcup G$ may provide the function from the subset of $\lambda$ to $({^\theta}\lambda)^M$, but I don't know how to proceed it. I would appreciate your help.
See this post where Monroe explained this.
Let $cf(\lambda) = \theta < \lambda$. I think the problem is missing the assumption $2^{< \lambda} > \lambda$.
It is easy to check that this forcing is $\theta$-closed (so it preserves all cardinals $\leq \theta$) and that it collapses $\lambda$ to $\theta$. As you noted, if $\lambda$ is a strong limit, this has $\lambda^{+}$-c.c. so it cannot collapse cardinals $\geq \lambda^{+}$. On the other hand, if $2^{\kappa} > \lambda$, for some $\kappa < \lambda$, then the generic adds a surjection from $\lambda$ to $2^{\kappa}$ so that $\lambda^{+}$ is collapsed to $\theta$ as well.
So $\lambda^+$ is collapsed iff $2^{< \lambda} > \lambda$.