simple uniform hashing: unclear definition of probability
First, the outcome of a situation being deterministic doesn't mean we will necessarily assign a probability of $1$ or $0$ to it. If I say I'm going to flip a coin, but it is actually a double-sided coin meaning both sides are heads or both sides are tails, the probability that it comes up heads is still $0.5$. Indeed, with respect to classical mechanics, the outcome of a normal coin flip is a deterministic function of the state of the universe, and even quantum mechanics would at best predict near certainty of the classical outcome. But you are correct, that if we knew $h$ and $k$ and $\ell$, then $h(k)=\ell$ would be either a logical tautology or a logical contradiction, and thus of probability $1$ or $0$ respectively.

We could consider any of $h$, $k$, or $\ell$ to be random variables, but the most natural choice would be $k$. I'll write $K$ for the random variable of the keys and reserve $k$ for some particular key. We have $$P(h(K)=\ell)=P(K\in h^{-1}(\ell)) = \mu_K(h^{-1}(\ell))$$ where $\mu_K$ is the probability measure corresponding to $K$. The simple uniform hashing assumption (SUHA) states that $$\mu_K(h^{-1}(\ell))=\frac{1}{m}$$ for every $\ell\in\{0,\dots,m-1\}$. This is simply a constraint on the probability measure $\mu_K$. There is no assumption that $K$ is "uniformly distributed" in any (other) sense. $K$ could be wildly "non-uniform" and still satisfy SUHA. Indeed, it would have to be for some hash functions, $h$.

For example, let's say $m=2$ and $h^{-1}(0)=\{k_0\}$ and thus $h^{-1}(1)=X\setminus \{k_0\}$ where $X$ is the domain of $h$. Let's further say $|X|=101$. Then for this $h$ to satisfy SUHA, it must be the case that $P(K=k_0)=\frac{1}{2}$. For $k\in X\setminus \{k_0\}$, it could be the case that there's a $k_1\in X\setminus \{k_0\}$ such that $P(K=k_1)=\frac{1}{2}$ and thus if $k\neq k_1$ then $P(K=k)=0$. Or maybe, $P(K=k)=\frac{1}{200}$ for any $k\in X\setminus \{k_0\}$. In fact, any distribution over the elements of $X\setminus \{k_0\}$ that sums to $\frac{1}{2}$ will suffice.

Typically, we would specify a distribution for $K$ and then SUHA would constrain what $h$ could be. The point is that the hash function is "well matched" to the distribution of keys whatever that is. There is no sense in stating a hash function satisfies SUHA without specifying the distribution of keys, but we can say a pair of a hash function and a distribution of keys satisfy SUHA while giving the details of neither. This is all that is needed to prove many results about hashing. We don't want to limit ourselves to only talking about "uniformly" distributed keys. A hash table will still gave the same performance guarantees for "non-uniformly" distributed keys as long as the hash function is chosen to "compensate" for the "non-uniform" distribution, i.e. to satisfy SUHA.