First of all, does $f(x,y)$ continuous in $\mathbb{R}^2$ imply it is continuous at a fixed $x$ as a function of $y$?
Yes, that's correct and holds for any continuous function $f: \mathbb{R}^2 \to \mathbb{R}$. For a fixed point $(x_0,y_0)$ continuity at $(x_0,y_0)$ means that for all $\epsilon>0$ there exists $\delta>0$ such that $$|f(x,y)-f(x_0,y_0)| \leq \epsilon \qquad \text{for all} \, \, |(x_0,y_0)-(x,y)| \leq \delta. \tag{1}$$
Now for fixed $x_0,y_0$ and $\epsilon>0$ choose $\delta>0$ as above, then we have $$|f(x_0,y)-f(x_0,y_0)| \leq \epsilon \qquad \text{for all} \, \, |y-y_0| \leq \delta$$ which shows that $f(x_0,\cdot)$ is continuous at $y=y_0$. Since $x_0,y_0$ are arbitrary, this finishes the proof.
If so does this imply $f$ is uniformly continuous on $[a,b]$ as a function of $y$?
Yeah, any continuous function which is defined on a compact interval is uniformly continuous, see this question; however, this is not needed for the proof of the assertion.
If so then $f$ is bounded by its supremum on the finite interval and hence its integral exists?
Correctly. (For this we don't need absolute continuity; continuity is enough.) Note that the continuity also implies measurability of $f(x,\cdot)$ for each $x$.
Do we turn this problem into a more tractable form by changing the derivative into a limit of sequence of functions $n(f(x+1/n,y)-f(x,y))$.
Yes, exactly. Set
$$F(x) := \int_{[a,b]} f(x,y) \, dy.$$
We have to show that $$\frac{\partial}{\partial x} F(x) = \int_{[a,b]} \partial_x f(x,y) \, dy. \tag{2}$$
First of all, note that $F$ is well-defined because $f(x,\cdot)$ is continuous (see above) and, moreover,
$$\frac{F(x+1/n)-F(x)}{1/n} = n \int_{[a,b]} (f(x+1/n,y)-f(x,y)) \, dy.$$
For fixed $x$ we define a sequence of auxiliary functions by $$u_n(y):= n (f(x+1/n,y)-f(x,y)).$$
Since $f(\cdot,y)$ is, by assumption, differentiable, we have
$$u_n(y) \xrightarrow[]{n \to \infty} \partial_x f(x,y).$$
On the other hand, the mean value theorem shows
$$|u_n(y)| \leq \sup_{\lambda \in [0,1]} |\partial_x f(x+\lambda/n,y)| \leq \sup_{y \in [a,b]} \sup_{|u-x| \leq 1} |\partial_x f(u,y)|.$$
Since $\partial_x f$ is continuous, the right-hand side is a (finite) constant and therefore integrable on the finite interval $[a,b]$. This means that we have found an integrable dominating function for $u_n$. Applying the dominated convergence theorem, we find that
$$\frac{F(x+1/n)-F(x)}{1/n} = \int_{[a,b]} u_n(y) \, dy \xrightarrow[]{n \to \infty} \int_{[a,b]} \partial_x f(x,y) \, dy.$$
This proves $(2)$.
