Concerning the finite sum, I am not sure you can get it (maybe for particular values of $ \nu $). Here is a possible "closed" expression for your sum. Start by writing $ \lambda := \mu + \alpha $ with $ \alpha > 0 $ and
$$ \frac{ (\mu)_k }{ (\lambda)_k } = \frac{ (\mu)_k }{ (\mu + \alpha)_k } = \mathbb{E}(\beta_{\mu, \alpha}^k) $$
where $ \beta_{\mu, \alpha} $ is a random variable Beta distributed (see wikipedia for the moments and the definition ; all I do is to write your quotient of Pochammers with a Beta integral). Up to this expectation, you are left with computing
$$ A := \sum_{k \geq 0} \frac{x^k }{k! } J_{k + \nu}(z) $$
Now, use the following representation (found on wikipedia) :
$$ J_\alpha(z) = \frac{ (z/2)^\alpha }{ \Gamma(\alpha + \frac{1}{2}) } \int_{ [-1, 1] } e^{i sz } (1 - s^2)^{ \alpha - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } } $$
valid for $ \alpha > \frac{1}{2} $ and $ z \in \mathbb{C} $ (you must hence suppose that your $ \nu $ is greater than $ \frac{1}{2} $ ; if not, you will have to adapt).

Using dominated convergence and exchange of integral and sum, you are left with
\begin{align*}%$
A & = \int_{ [-1, 1] } (z/2)^\nu \sum_{k \geq 0} \frac{ (x (1 - s^2) z/2 )^k }{k! \Gamma(\nu + k + \frac{1}{2}) } e^{i sz } (1 - s^2)^{ \nu - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } } \\
& = \int_{ [-1, 1] } (2 x (s^2 - 1) )^{-\nu} J_{\nu + 1/2} (x (s^2 - 1 ) z ) e^{i sz } (1 - s^2)^{ \nu - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } }
\end{align*}
where I use the definition of the Bessel function as a series (here again, see wikipedia).

You can then use the previous integral representation of the Bessel function to conclude (remember that $ x = \beta_{\mu, \alpha} t z/2 $ and you have to take the expectation in $ \beta_{\mu, \alpha} $). Your representation is then a triple integral involving classical functions.