Time reversal of input signals in convolution
This question appears to involve confusion about the relationship between convolution and cross-correlation. To illustrate the relationship, consider the two rectangular pulses shown below:

The shorter pulse has width = 1 and the longer pulse has width = 2. For simplicity, assume unity heights for the pulses.
In my answer to a previous question, I used a figure from a textbook 1 to illustrate the convolution of $\mathrm {x_1(\lambda)}$ and $\mathrm {x_2(\lambda)}$, denoted by $\mathrm {x_1*x_2}$.
The convolution integral was given by
$$\mathrm {x_1*x_2 =\int_{-\infty}^{\infty}x_1(\lambda)x_2(t-\lambda)\,\mathrm d\lambda} \tag{1}$$
so $\mathrm {x_2(\lambda)}$ was reflected around the ordinate and is the function that gets shifted. This is clearly shown in the figure in my linked answer.
But it is also possible to reflect $\mathrm {x_1(\lambda)}$ about the ordinate and shift it. Then the convolution would be denoted by $\mathrm {x_2*x_1}$ and the convolution integral would be
$$\mathrm {x_2*x_1 =\int_{-\infty}^{\infty}x_1(t-\lambda)x_2(\lambda)\,\mathrm d\lambda} \tag{2}$$
This is illustrated the following figure that I drew in the same fashion as the figure in my linked answer:

Thus $\mathrm {x_1*x_2 = x_2*x_1}$, i.e., it does not matter which function is reversed for the convolution. Of course, if a function is symmetric about its center, then reversing becomes rather trivial. In this example, both rectangular pulses are symmetric about their centers, so the situation is especially simple.
In corroboration of the above, the following (through their equation 1.5-3) is directly quoted from Blinchikoff and Zverev [2, p. 12]:

Let p(t) and q(t) each be a piecewise continuous function for $-\infty &lt; t &lt; \infty$. The convolution of p and q, denoted by $p*q$, is defined to be the third function
$$\mathrm {v(t) = p*q =\int_{-\infty}^{\infty}p(t-\tau)g(\tau)\,\mathrm d\tau} \tag{1.5-1}$$
and the integral on the right is called the convolution integral. Furthermore, the convolution of q and p, $q*p$, is identical to p*q, that is,
$$q*p = p*q \tag{1.5-2}$$
This is easily shown by letting $\mathrm {x = t -\tau}$, $\mathrm {dx = -d\tau}$ in (1.5-1), yielding
$$\mathrm {p*q =\int_{-\infty}^{\infty}p(x)g(t-x)\,\mathrm dx} = q*p \tag{1.5-3}$$

Therefore, in answer to the OP's second question, it does not matter mathematically if f(t) or g(t) is the reversed function, but the calculation may be simplified, in practice, if either function is symmetric about its center.
For cross-correlation, all that is necessary is to shift one function by $\tau$, as per this photo image from [2, p. 305]:

So cross-correlation calculations are quite similar to convolution calculations, but note the important considerations discussed directly below Blinchikoff and Zverev's Table 7-2. I suggest that the OP think about their first question, i.e., what happens when both f(t) and g(t) are reversed and the overlap integrals are calculated.
1  C.D. McGillem, G.R. Cooper, &quot;Continuous and Discrete Signal and System Analysis&quot;, 2nd Ed., Holt, Rinehart and Winston, NY, ©1984, pp. 58-59.
2  H.J. Blinchikoff, A.I. Zverev, &quot;Filtering in the Time and Frequency Domains&quot;, Wiley-Interscience, John Wiley &amp; Sons, NY, ©1976.