I think your proof has the right idea, but I would advise against using indefinite integrals in a formal proof - you would be better off looking at a specific antiderivative. Namely, assume $P$ is defined on $[a,b]\times[c,d]$ and let
$$Q(x,y):=\int_a^xP(t,y)\,dt,$$
so $\frac{\partial Q}{\partial x}(x,y)=P(x,y)$. Your question becomes about whether we have
$$\frac{\partial Q}{\partial y}(x,y)=\int_a^x\frac{\partial P}{\partial y}(t,y)\,dt.$$
This is, of course, a question about interchanging limits, and as in your proof, the bounded convergence theorem is the way to go. We have
$$\frac{Q(x,y+h)-Q(x,y)}{h}=\int_a^xR_h(t,y)\,dt$$
where $R_h(t,y)=\frac{P(t,y+h)-P(t,y)}h$. Clearly $R_h(t,y)\to\frac{\partial P}{\partial y}(t,y)$ as $h\to0$. Moreover, by the intermediate value theorem, $R_h(t,y)=\frac{\partial P}{\partial y}(t,c)$ for some $c$ between $y$ and $y+h$. Since $P$ is $C^1$, $\frac{\partial P}{\partial y}$ is bounded. This is enough to apply the bounded convergence theorem and conclude.
In terms of Riemann vs Lebesgue integral, I would say as a general rule you should always assume you're working with the Lebesgue integral (once you know it) except for very specific circumstances, such as discussing the convergence of an improper integral which is not absolutely convergent. It is simple to show that if $f$ is Riemann integrable (on a bounded interval) then it is also Lebesgue integrable, and we have many more tools available with Lebesgue integration, so we may as well use them.
You need to be careful with any notation since often authors define and use symbols in different manner.
Although in this particular case, my mathematical experience tells me that most common are:
1. "Gradient's variable":
following Ruben Tobar's answer, it may be notation providing information with respect to which variables the whole gradient should be taken. To be more precise: if $x_i\in \mathbb{R}$ for $i= 1,...,n$ and $\mathbf{x}\in\mathbb{R}^n$ is a vector defined as $\mathbf{x}=\{x_1,...,x_n\}$, given differentiable $f(\mathbf{x}):\mathbb{R}^n\rightarrow\mathbb{R}$ we have:
$$\nabla_{\mathbf{x}}f=(\partial_{x_1}f,...,\partial_{x_n}f)$$
Therefore $\nabla_{\mathbf{x}}f$ is a vector whose entities $\partial_{x_i}f$ are just standard derivatives of multi-variable function $f$ with respect to the real variable $x_i$ (so called partial derivatives). Note that I did not write here explicitly dependence of $f$ on $\mathbf{x}$ since it would be redundant and erroneous, but nevertheless both left hand side and all the entities on the right hand side are functions dependent on $\mathbf{x}$.
To elaborate about the notation a little: it is quite commonly used in partial differential equations, where you have multiple different variables and not to confuse your reader, when you use (or define) a differential operator you stress the variables used in definition of this operator. For example consider the following: take $\mathbf{x}\in{\mathbb{R}^n}$ defined as above and similarly $\mathbf{z}\in{\mathbb{R}^d}$ for some natural $r,d>1$ and $t\in\mathbb{R}$. Problem is to find a function $f(t,\mathbf{x},\mathbf{z}):\mathbb{R}\times \mathbb{R}^n\times\mathbb{R}^d\rightarrow\mathbb{R}$ solving differential equation. Without stressing the variables the following differential equation is ambiguous and difficult to write in any different way:
$$f_t + \nabla_\mathbf{x}f +{\rm div}_\mathbf{z}f = 0$$
where ${\rm div}_\mathbf{z}$ is different differential operator (called divergence) defined as $\sum_i\partial_{z_i}$, $i=1,...d.$
2. Directional derivative:
As you see in https://en.wikipedia.org/wiki/Directional_derivative (where you can find more information) it is a very common notation for directional derivative.
Not in full generality: Given any $f(\mathbf{x}):\mathbb{R}^n\rightarrow\mathbb{R}$ and a vector $\mathbf{v}\in \mathbb{R}^n$:
$$\nabla_{\mathbf{v}} f(\mathbf{x}):=\nabla f(\mathbf{x})\cdot\mathbf{v},$$
where on the right hand side we have standard scalar product.
3. Gradient at x?:
In my opinion (some may disagree): it is neither common, nor useful notation. Although, using our notation it would mean:
$$\nabla_\mathbf{x}f:=[\nabla f](\mathbf{x}):=(\partial_{x_1} f(\mathbf{x}),...,\partial_{x_n} f(\mathbf{x}))$$
Where $\mathbf{x}\in\mathbb{R}^n$ is fixed and partial derivatives of $f$ are evaluated at $\mathbf{x}$. Note that unlike in point 1. where we didn't stress dependence on the variable, here we write explicitly that functions are evaluated at $\mathbf{x}.$
Best Answer
The directional derivative of $f$, at $x$, in direction $u$ is defined like so:
$$D_u f(x) = \lim_{t \to 0} \frac{f(x + tu) - f(x)}{t}.$$
This can exist for very non-smooth functions. For example, when $f$ is the indicator function for the $x$-axis (i.e. takes the value $1$ on $\mathbb{R} \times \{ 0 \}$ and $0$ elsewhere), then the directional derivative at $0$ exists in the directions $(1, 0)$ and $(-1, 0)$ (and is equal to $0$). Note that the $y$-partial derivative doesn't exist in this case, and the dot product formula is nonsense.
If $f$ is differentiable in the multivariate sense (one sufficient condition is that the partial derivatives exist in a neighbourhood of the point and are continuous at the point), then the dot product formula holds. In fact, the existence of a vector $v$ such that, for all direction vectors $u$,
$$v \cdot u = D_u f(x)$$
is practically the definition of $f$ being differentiable at the point $x$. Such a vector must necessarily be $\nabla f (x)$.