Purpose of hashing last block in Merkle-Damgård? Is hashing the last block in the Merkle-Damgård necessary in preventing collisions? i.e.  What if I just outputted $z_B || L$, where $z_i$ is the hash of the last block of the message, L is the length of the message and B is the number of blocks, instead of $H(z_B || L)$?
The purpose of hashing the last block is simply to prevent leaking information like, $L$, the length of the input. It's also important so that if $L\leq \langle 2 \rangle$, then $H$ is still compressing the input. 

That being said, if you don't hash the last block of the Merkle-Damgård transform, $H$ is still collision-resistant, assuming that $h$ is collision resistant.

Let $\Pi=(Gen,H)$ and $\Pi'=(Gen,h)$. Assume $h$ is a collision-resistant hash function. Suppose for a contradiction that we can find a poly-time adversary $\mathcal{A}$ that finds a collision in $H$ with non-negligible probability $\delta(n)$. We will now construct a poly-time adversary $\mathcal{A}'$ that will find a collision in $h$.

$\mathcal{A}'$ begins by running $\mathcal{A}$ and finding a collision in $H$ such that $H^s(x)=x_B\Vert L=x_B'\Vert L'=H^s(x')$ and $x\neq x'$. Since the length is sent in the open and $x_B\Vert L=x_B'\Vert L'$, then it must be the case that $L=L'$. Define $\mathcal{A}'$ to output whatever $\mathcal{A}$ outputs.

Because $|x|=|x'|=L$, then we can find a collision by proceeding inductively. Since $x\neq x'$, then there must exist some $i$ where $1\leq i\leq B$ such that $x_i\Vert z_{i-1} \neq x_i'\Vert z_{i-1}'$.

Working backwards from the output, find the first such $i$. It must be the case that $h^s(x_i\Vert z_{i-1}) = h^s(x_i'\Vert z_{i-1}')$ and so we have found a collision in $h$. Have $\mathcal{A}$ output $x_i\Vert z_{i-1}$ and $x_i'\Vert z_{i-1}'$.

But this means that we have found a collision in $h$ by calling $\mathcal{A}$ only a polynomial number of times which implies that $\mathcal{A}'$ also runs in polynomial time i.e.
$$\text{Pr[Hash-Coll}_{\mathcal{A}',\Pi'}(n)=1]$$
is not bounded by any negligible function, contradicting the collision-resistance of $h$.